In this collection we present 20 practice problems, worked through step by step, on the Stolz–Cesàro theorem and its principal corollaries. The aim is to learn how to recognise when the theorem may be applied and how to use it correctly in evaluating limits of ratios of sequences.
In the problems we shall check carefully the hypotheses of the result used each time. When we apply the Stolz–Cesàro theorem directly, the sequence in the denominator must be eventually positive, strictly increasing, and divergent to \(+\infty\). When we use one of its corollaries, we shall instead verify the corresponding assumptions, such as positivity of the terms and the existence of the relevant limit.
The problems are arranged in order of increasing difficulty: the first few serve to fix the method, while the later ones call for greater care in the choice of sequences and in evaluating the limit of the increments.
Exercise 1 — level ★☆☆☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{1+2+\cdots+n}{n^2}. \]
Answer
\[ \lim_{n\to+\infty}\frac{1+2+\cdots+n}{n^2}=\frac{1}{2}. \]
Solution
Set
\[ a_n=1+2+\cdots+n, \qquad b_n=n^2. \]
We wish to evaluate the limit of the ratio
\[ \frac{a_n}{b_n}. \]
For \(n\ge 1\) we have \(b_n=n^2>0\). Moreover, \(b_n\) is strictly increasing and
\[ \lim_{n\to+\infty}b_n=+\infty. \]
We may therefore apply the Stolz–Cesàro theorem, provided the limit of the ratio of increments exists.
We compute:
\[ a_{n+1}-a_n=(1+2+\cdots+n+(n+1))-(1+2+\cdots+n)=n+1, \]
while
\[ b_{n+1}-b_n=(n+1)^2-n^2=2n+1. \]
Hence
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{n+1}{2n+1}. \]
We evaluate the limit:
\[ \lim_{n\to+\infty}\frac{n+1}{2n+1} = \lim_{n\to+\infty}\frac{1+\displaystyle\frac{1}{n}}{2+\displaystyle\frac{1}{n}} = \frac{1}{2}. \]
By the Stolz–Cesàro theorem it follows that
\[ \lim_{n\to+\infty}\frac{a_n}{b_n} = \frac{1}{2}. \]
Therefore
\[ \lim_{n\to+\infty}\frac{1+2+\cdots+n}{n^2} = \frac{1}{2}. \]
Exercise 2 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{1^2+2^2+\cdots+n^2}{n^3}. \]
Answer
\[ \lim_{n\to+\infty}\frac{1^2+2^2+\cdots+n^2}{n^3}=\frac{1}{3}. \]
Solution
Set
\[ a_n=1^2+2^2+\cdots+n^2, \qquad b_n=n^3. \]
For \(n\ge 1\) we have \(b_n=n^3>0\). Moreover, \(b_n\) is strictly increasing and tends to \(+\infty\).
The hypotheses on the denominator are therefore satisfied.
We now compute the increments. For the sequence in the numerator we have
\[ a_{n+1}-a_n=(n+1)^2. \]
For the sequence in the denominator:
\[ b_{n+1}-b_n=(n+1)^3-n^3. \]
Expanding the cube:
\[ (n+1)^3=n^3+3n^2+3n+1. \]
Hence
\[ b_{n+1}-b_n=3n^2+3n+1. \]
The ratio of increments is
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{(n+1)^2}{3n^2+3n+1}. \]
We evaluate the limit by dividing numerator and denominator by \(n^2\):
\[ \lim_{n\to+\infty}\frac{(n+1)^2}{3n^2+3n+1} = \lim_{n\to+\infty} \frac{1+\displaystyle\frac{2}{n}+\displaystyle\frac{1}{n^2}}{3+\displaystyle\frac{3}{n}+\displaystyle\frac{1}{n^2}} = \frac{1}{3}. \]
By the Stolz–Cesàro theorem we obtain
\[ \lim_{n\to+\infty}\frac{1^2+2^2+\cdots+n^2}{n^3} = \frac{1}{3}. \]
Exercise 3 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{1+\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}+\cdots+\displaystyle\frac{1}{n}}{n}. \]
Answer
\[ \lim_{n\to+\infty}\frac{1+\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}+\cdots+\displaystyle\frac{1}{n}}{n}=0. \]
Solution
Set
\[ a_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}, \qquad b_n=n. \]
The denominator \(b_n=n\) is positive for \(n\ge 1\), is strictly increasing, and tends to \(+\infty\).
We may therefore apply the Stolz–Cesàro theorem.
We compute the increments:
\[ a_{n+1}-a_n=\frac{1}{n+1}, \]
while
\[ b_{n+1}-b_n=(n+1)-n=1. \]
Hence
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{1}{n+1}. \]
Since
\[ \lim_{n\to+\infty}\frac{1}{n+1}=0, \]
by the Stolz–Cesàro theorem it follows that
\[ \lim_{n\to+\infty}\frac{a_n}{b_n}=0. \]
Therefore
\[ \lim_{n\to+\infty}\frac{1+\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}+\cdots+\displaystyle\frac{1}{n}}{n}=0. \]
This example shows that the sequence of partial sums of the harmonic series grows much more slowly than \(n\).
Exercise 4 — level ★★★☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{\log(n!)}{n\log n}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\log(n!)}{n\log n}=1. \]
Solution
Recall that
\[ \log(n!)=\log(1\cdot 2\cdot 3\cdots n). \]
Set
\[ a_n=\log(n!), \qquad b_n=n\log n. \]
We consider the ratio for \(n\ge 2\), so that \(b_n>0\).
Moreover, \(b_n=n\log n\) is eventually strictly increasing. Indeed, the function \(x\log x\) has derivative \(1+\log x\), which is positive for \(x>e^{-1}\). In particular, for \(n\ge 1\), the sequence \(n\log n\) is strictly increasing.
Finally,
\[ \lim_{n\to+\infty}n\log n=+\infty. \]
We may therefore use the Stolz–Cesàro theorem.
We compute the increments. For the numerator:
\[ a_{n+1}-a_n = \log((n+1)!)-\log(n!) = \log(n+1). \]
For the denominator:
\[ b_{n+1}-b_n = (n+1)\log(n+1)-n\log n. \]
We rewrite the denominator in a more convenient form:
\[ (n+1)\log(n+1)-n\log n = \log(n+1)+n\log\left(\displaystyle\frac{n+1}{n}\right). \]
Hence
\[ b_{n+1}-b_n = \log(n+1)+n\log\left(1+\displaystyle\frac{1}{n}\right). \]
The ratio of increments is then
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{\log(n+1)} {\log(n+1)+n\log\left(1+\displaystyle\frac{1}{n}\right)}. \]
Dividing numerator and denominator by \(\log(n+1)\):
\[ \frac{\log(n+1)} {\log(n+1)+n\log\left(1+\displaystyle\frac{1}{n}\right)} = \frac{1} {1+\displaystyle\frac{n\log\left(1+\displaystyle\frac{1}{n}\right)}{\log(n+1)}}. \]
We know that
\[ n\log\left(1+\displaystyle\frac{1}{n}\right)\to 1, \]
while
\[ \log(n+1)\to+\infty. \]
Hence
\[ \frac{n\log\left(1+\displaystyle\frac{1}{n}\right)}{\log(n+1)}\to 0. \]
It follows that
\[ \lim_{n\to+\infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{1}{1+0} = 1. \]
By the Stolz–Cesàro theorem we conclude that
\[ \lim_{n\to+\infty}\frac{\log(n!)}{n\log n}=1. \]
Exercise 5 — level ★★★★☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}}{n^{3/2}}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}}{n^{3/2}}=\frac{2}{3}. \]
Solution
Set
\[ a_n=\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}, \qquad b_n=n^{3/2}. \]
For \(n\ge 1\) we have \(b_n>0\). Moreover, \(b_n=n^{3/2}\) is strictly increasing and
\[ \lim_{n\to+\infty}n^{3/2}=+\infty. \]
The hypotheses of the Stolz–Cesàro theorem concerning the denominator are therefore satisfied.
We compute the increments:
\[ a_{n+1}-a_n=\sqrt{n+1}. \]
Moreover,
\[ b_{n+1}-b_n=(n+1)^{3/2}-n^{3/2}. \]
The ratio of increments is therefore
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{\sqrt{n+1}}{(n+1)^{3/2}-n^{3/2}}. \]
We factor out \(\sqrt{n+1}\) in the denominator. Since
\[ (n+1)^{3/2}=(n+1)\sqrt{n+1} \]
and
\[ n^{3/2}=n\sqrt n=n\sqrt{n+1}\sqrt{\displaystyle\frac{n}{n+1}}, \]
we obtain
\[ (n+1)^{3/2}-n^{3/2}=\sqrt{n+1}\left((n+1)-n\sqrt{\displaystyle\frac{n}{n+1}}\right). \]
Hence
\[ \frac{\sqrt{n+1}}{(n+1)^{3/2}-n^{3/2}}=\frac{\sqrt{n+1}}{\sqrt{n+1}\left((n+1)-n\sqrt{\displaystyle\frac{n}{n+1}}\right)}=\frac{1}{(n+1)-n\sqrt{\displaystyle\frac{n}{n+1}}}. \]
We therefore examine the denominator
\[ (n+1)-n\sqrt{\frac{n}{n+1}}. \]
Set
\[ t_n=\frac{1}{n+1}. \]
Then \(t_n\to 0^+\) and
\[ \frac{n}{n+1}=1-t_n. \]
Moreover, \(n+1=\displaystyle \frac{1}{t_n}\) and \(n=\displaystyle \frac{1-t_n}{t_n}\). Hence
\[ (n+1)-n\sqrt{\frac{n}{n+1}} = \frac{1}{t_n}-\frac{1-t_n}{t_n}\sqrt{1-t_n}. \]
Factoring out \(1/t_n\), we obtain
\[ (n+1)-n\sqrt{\frac{n}{n+1}} = \frac{1-(1-t_n)^{3/2}}{t_n}. \]
Using the fundamental limit, equivalent to the differentiability of the function \(x^{3/2}\) at \(x=1\),
\[ \lim_{t\to 0^+}\frac{1-(1-t)^{3/2}}{t}=\frac{3}{2}. \]
Hence
\[ (n+1)-n\sqrt{\frac{n}{n+1}}\to\frac{3}{2}. \]
Consequently
\[ \lim_{n\to+\infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{1}{\displaystyle \frac{3}{2}} = \frac{2}{3}. \]
By the Stolz–Cesàro theorem it follows that
\[ \lim_{n\to+\infty}\frac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}}{n^{3/2}} = \frac{2}{3}. \]
Exercise 6 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{1^3+2^3+\cdots+n^3}{n^4}. \]
Answer
\[ \lim_{n\to+\infty}\frac{1^3+2^3+\cdots+n^3}{n^4}=\frac{1}{4}. \]
Solution
Set
\[ a_n=1^3+2^3+\cdots+n^3, \qquad b_n=n^4. \]
For \(n\ge 1\) we have \(b_n=n^4>0\). Moreover, \(b_n\) is strictly increasing and
\[ \lim_{n\to+\infty}b_n=+\infty. \]
The hypotheses on the denominator required by the Stolz–Cesàro theorem are therefore satisfied.
We compute the increments. For the sequence in the numerator:
\[ a_{n+1}-a_n=(n+1)^3. \]
For the sequence in the denominator:
\[ b_{n+1}-b_n=(n+1)^4-n^4. \]
Expanding:
\[ (n+1)^4=n^4+4n^3+6n^2+4n+1. \]
Hence
\[ b_{n+1}-b_n=4n^3+6n^2+4n+1. \]
The ratio of increments is
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{(n+1)^3}{4n^3+6n^2+4n+1}. \]
Dividing numerator and denominator by \(n^3\):
\[ \frac{(n+1)^3}{4n^3+6n^2+4n+1} = \frac{1+\displaystyle\frac{3}{n}+\displaystyle\frac{3}{n^2}+\displaystyle\frac{1}{n^3}} {4+\displaystyle\frac{6}{n}+\displaystyle\frac{4}{n^2}+\displaystyle\frac{1}{n^3}}. \]
Hence
\[ \lim_{n\to+\infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{1}{4}. \]
By the Stolz–Cesàro theorem it follows that
\[ \lim_{n\to+\infty}\frac{1^3+2^3+\cdots+n^3}{n^4} = \frac{1}{4}. \]
Exercise 7 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{k}{k+1}. \]
Answer
\[ \lim_{n\to+\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{k}{k+1}=1. \]
Solution
We write the limit in the form of a ratio:
\[ \frac{1}{n}\sum_{k=1}^{n}\frac{k}{k+1} = \frac{\displaystyle\sum_{k=1}^{n}\frac{k}{k+1}}{n}. \]
Set
\[ a_n=\sum_{k=1}^{n}\frac{k}{k+1}, \qquad b_n=n. \]
The denominator \(b_n=n\) is positive for \(n\ge 1\), is strictly increasing, and tends to \(+\infty\).
We may therefore apply the Stolz–Cesàro theorem.
We compute the increments. For the numerator:
\[ a_{n+1}-a_n=\frac{n+1}{n+2}. \]
For the denominator:
\[ b_{n+1}-b_n=1. \]
Hence
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{n+1}{n+2}. \]
We evaluate the limit:
\[ \lim_{n\to+\infty}\frac{n+1}{n+2} = \lim_{n\to+\infty} \frac{1+\displaystyle\frac{1}{n}}{1+\displaystyle\frac{2}{n}} = 1. \]
By the Stolz–Cesàro theorem we obtain
\[ \lim_{n\to+\infty} \frac{\displaystyle\sum_{k=1}^{n}\frac{k}{k+1}}{n} = 1. \]
Hence
\[ \lim_{n\to+\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{k}{k+1}=1. \]
This result is also consistent with Cesàro's theorem, since the sequence
\[ \frac{k}{k+1} \]
tends to \(1\), and hence its arithmetic mean tends to \(1\) as well.
Exercise 8 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}. \]
Answer
\[ \lim_{n\to+\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=0. \]
Solution
We rewrite the limit as a ratio:
\[ \frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{k}} = \frac{\displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}}{n}. \]
Set
\[ a_n=\sum_{k=1}^{n}\frac{1}{\sqrt{k}}, \qquad b_n=n. \]
The sequence \(b_n=n\) is positive for \(n\ge 1\), strictly increasing, and divergent to \(+\infty\).
We may therefore apply the Stolz–Cesàro theorem.
We compute the increments:
\[ a_{n+1}-a_n=\frac{1}{\sqrt{n+1}}, \qquad b_{n+1}-b_n=1. \]
Hence
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{1}{\sqrt{n+1}}. \]
Since
\[ \lim_{n\to+\infty}\frac{1}{\sqrt{n+1}}=0, \]
the Stolz–Cesàro theorem implies
\[ \lim_{n\to+\infty} \frac{\displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}}{n} = 0. \]
Therefore
\[ \lim_{n\to+\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=0. \]
Here too the result is consistent with Cesàro's theorem, since
\[ \frac{1}{\sqrt{k}}\to 0. \]
Exercise 9 — level ★★★☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\sqrt[n]{\prod_{k=1}^{n}\left(1+\frac{1}{k}\right)}. \]
Answer
\[ \lim_{n\to+\infty}\sqrt[n]{\prod_{k=1}^{n}\left(1+\frac{1}{k}\right)}=1. \]
Solution
Consider the sequence
\[ c_k=1+\frac{1}{k}, \qquad k\ge 1. \]
For every \(k\ge 1\) we have \(c_k>0\). Moreover,
\[ \lim_{k\to+\infty}c_k= \lim_{k\to+\infty}\left(1+\frac{1}{k}\right)=1. \]
We wish to evaluate the limit of the geometric mean of the first \(n\) terms of the sequence \(c_k\):
\[ \sqrt[n]{\prod_{k=1}^{n}c_k} = \sqrt[n]{\prod_{k=1}^{n}\left(1+\frac{1}{k}\right)}. \]
Since \(c_k>0\) and \(c_k\to 1\), we may apply the corollary concerning the geometric mean.
This corollary states that if a sequence of positive terms converges to a positive limit \(L\), then the geometric mean of its initial terms likewise converges to \(L\).
In the present case \(L=1\). Hence
\[ \lim_{n\to+\infty}\sqrt[n]{\prod_{k=1}^{n}\left(1+\frac{1}{k}\right)} = 1. \]
We observe that the result does not depend on the fact that the product
\[ \prod_{k=1}^{n}\left(1+\frac{1}{k}\right) \]
grows with \(n\). What is decisive is that we are taking the \(n\)-th root of the product, that is, a geometric mean.
Exercise 10 — level ★★★☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\sqrt[n]{5^n\frac{2n+1}{n+3}}. \]
Answer
\[ \lim_{n\to+\infty}\sqrt[n]{5^n\frac{2n+1}{n+3}}=5. \]
Solution
Set
\[ a_n=5^n\frac{2n+1}{n+3}. \]
For every \(n\ge 0\) we have \(a_n>0\). We may therefore use the corollary concerning the limit of the \(n\)-th root.
We compute the ratio of two consecutive terms:
\[ \frac{a_{n+1}}{a_n} = \frac{5^{n+1}\displaystyle\frac{2n+3}{n+4}} {5^n\displaystyle\frac{2n+1}{n+3}}. \]
Simplifying \(5^n\), we obtain
\[ \frac{a_{n+1}}{a_n} = 5\cdot \frac{2n+3}{n+4} \cdot \frac{n+3}{2n+1}. \]
We evaluate the limit:
\[ \lim_{n\to+\infty} 5\cdot \frac{2n+3}{n+4} \cdot \frac{n+3}{2n+1} = 5\cdot 2\cdot\frac{1}{2} = 5. \]
Hence
\[ \lim_{n\to+\infty}\frac{a_{n+1}}{a_n}=5. \]
Since \(a_n>0\) and the limit of the ratio of consecutive terms exists and equals \(5>0\), the Stolz–Cesàro corollary on the \(n\)-th root yields
\[ \lim_{n\to+\infty}\sqrt[n]{a_n}=5. \]
Substituting the definition of \(a_n\), we obtain
\[ \lim_{n\to+\infty}\sqrt[n]{5^n\frac{2n+1}{n+3}}=5. \]
Exercise 11 — level ★★★☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{\sum_{k=1}^{n}k\log k}{n^2\log n}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\sum_{k=1}^{n}k\log k}{n^2\log n}=\frac{1}{2}. \]
Solution
Set
\[ a_n=\sum_{k=1}^{n}k\log k, \qquad b_n=n^2\log n. \]
We consider the ratio for \(n\ge 2\), so that \(b_n>0\). Moreover, \(b_n=n^2\log n\) is eventually strictly increasing and
\[ \lim_{n\to+\infty}b_n=+\infty. \]
We may therefore apply the Stolz–Cesàro theorem.
We compute the increments. For the numerator we have
\[ a_{n+1}-a_n=(n+1)\log(n+1). \]
For the denominator:
\[ b_{n+1}-b_n=(n+1)^2\log(n+1)-n^2\log n. \]
We rewrite this difference in a form better suited to the limit:
\[ b_{n+1}-b_n= \bigl((n+1)^2-n^2\bigr)\log(n+1)+n^2\bigl(\log(n+1)-\log n\bigr). \]
Since
\[ (n+1)^2-n^2=2n+1 \]
and
\[ \log(n+1)-\log n=\log\left(1+\frac{1}{n}\right), \]
we obtain
\[ b_{n+1}-b_n=(2n+1)\log(n+1)+n^2\log\left(1+\frac{1}{n}\right). \]
The ratio of increments is therefore
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{(n+1)\log(n+1)} {(2n+1)\log(n+1)+n^2\log\left(1+\displaystyle\frac{1}{n}\right)}. \]
Dividing numerator and denominator by \(n\log(n+1)\):
\[ \frac{(n+1)\log(n+1)} {(2n+1)\log(n+1)+n^2\log\left(1+\displaystyle\frac{1}{n}\right)} = \frac{1+\displaystyle\frac{1}{n}} {2+\displaystyle\frac{1}{n}+ \displaystyle\frac{n\log\left(1+\displaystyle\frac{1}{n}\right)}{\log(n+1)}}. \]
We know that
\[ n\log\left(1+\frac{1}{n}\right)\to 1, \]
while
\[ \log(n+1)\to+\infty. \]
Hence
\[ \frac{n\log\left(1+\displaystyle\frac{1}{n}\right)}{\log(n+1)}\to 0. \]
Hence
\[ \lim_{n\to+\infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{1}{2}. \]
By the Stolz–Cesàro theorem it follows that
\[ \lim_{n\to+\infty}\frac{\sum_{k=1}^{n}k\log k}{n^2\log n} = \frac{1}{2}. \]
Exercise 12 — level ★★★☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{2k^2-k+1}{k^2+1}. \]
Answer
\[ \lim_{n\to+\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{2k^2-k+1}{k^2+1}=2. \]
Solution
We rewrite the limit as a ratio:
\[ \frac{1}{n}\sum_{k=1}^{n}\frac{2k^2-k+1}{k^2+1} = \frac{\displaystyle\sum_{k=1}^{n}\frac{2k^2-k+1}{k^2+1}}{n}. \]
Set
\[ a_n=\sum_{k=1}^{n}\frac{2k^2-k+1}{k^2+1}, \qquad b_n=n. \]
The sequence \(b_n=n\) is positive for \(n\ge 1\), strictly increasing, and divergent to \(+\infty\).
We may therefore apply the Stolz–Cesàro theorem.
We compute the increments:
\[ a_{n+1}-a_n= \frac{2(n+1)^2-(n+1)+1}{(n+1)^2+1}, \qquad b_{n+1}-b_n=1. \]
Hence
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{2(n+1)^2-(n+1)+1}{(n+1)^2+1}. \]
Expanding the numerator:
\[ 2(n+1)^2-(n+1)+1 = 2n^2+3n+2. \]
Moreover,
\[ (n+1)^2+1=n^2+2n+2. \]
Hence
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{2n^2+3n+2}{n^2+2n+2}. \]
Dividing numerator and denominator by \(n^2\), we obtain
\[ \frac{2n^2+3n+2}{n^2+2n+2} = \frac{2+\displaystyle\frac{3}{n}+\displaystyle\frac{2}{n^2}} {1+\displaystyle\frac{2}{n}+\displaystyle\frac{2}{n^2}}. \]
Hence
\[ \lim_{n\to+\infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = 2. \]
By the Stolz–Cesàro theorem it follows that
\[ \lim_{n\to+\infty} \frac{\displaystyle\sum_{k=1}^{n}\frac{2k^2-k+1}{k^2+1}}{n} = 2. \]
Therefore
\[ \lim_{n\to+\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{2k^2-k+1}{k^2+1}=2. \]
The result is also consistent with Cesàro's theorem, since
\[ \frac{2k^2-k+1}{k^2+1}\to 2. \]
Exercise 13 — level ★★★☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\sqrt[n]{\prod_{k=1}^{n}\frac{3k+1}{k+2}}. \]
Answer
\[ \lim_{n\to+\infty}\sqrt[n]{\prod_{k=1}^{n}\frac{3k+1}{k+2}}=3. \]
Solution
Consider the sequence
\[ c_k=\frac{3k+1}{k+2}. \]
For every \(k\ge 1\) we have \(c_k>0\), since both numerator and denominator are positive.
Moreover,
\[ \lim_{k\to+\infty}c_k = \lim_{k\to+\infty}\frac{3k+1}{k+2} = \lim_{k\to+\infty} \frac{3+\displaystyle\frac{1}{k}}{1+\displaystyle\frac{2}{k}} = 3. \]
The required limit is the geometric mean of the first \(n\) terms of the sequence \(c_k\):
\[ \sqrt[n]{\prod_{k=1}^{n}c_k} = \sqrt[n]{\prod_{k=1}^{n}\frac{3k+1}{k+2}}. \]
Since \(c_k>0\) and \(c_k\to 3\), we may apply the corollary concerning the geometric mean.
This corollary states that if a sequence of positive terms converges to a positive limit \(L\), then the geometric mean of its initial terms converges to the same limit \(L\).
In the present case \(L=3\). Hence
\[ \lim_{n\to+\infty}\sqrt[n]{\prod_{k=1}^{n}\frac{3k+1}{k+2}}=3. \]
Exercise 14 — level ★★★★☆
Evaluate the limit
\[ \lim_{n\to+\infty}\sqrt[n]{\frac{(2n)!}{(n!)^2}}. \]
Answer
\[ \lim_{n\to+\infty}\sqrt[n]{\frac{(2n)!}{(n!)^2}}=4. \]
Solution
Set
\[ a_n=\frac{(2n)!}{(n!)^2}. \]
For every \(n\ge 1\) we have \(a_n>0\). We may therefore attempt to apply the corollary concerning the \(n\)-th root.
We compute the ratio of consecutive terms:
\[ \frac{a_{n+1}}{a_n} = \frac{\displaystyle\frac{(2n+2)!}{((n+1)!)^2}} {\displaystyle\frac{(2n)!}{(n!)^2}}. \]
Dividing by a fraction amounts to multiplying by its reciprocal, hence
\[ \frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{((n+1)!)^2} \cdot \frac{(n!)^2}{(2n)!}. \]
We now use the identities
\[ (2n+2)!=(2n+2)(2n+1)(2n)! \]
and
\[ (n+1)!=(n+1)n!. \]
Hence
\[ ((n+1)!)^2=(n+1)^2(n!)^2. \]
Substituting, we obtain
\[ \frac{a_{n+1}}{a_n} = \frac{(2n+2)(2n+1)(2n)!}{(n+1)^2(n!)^2} \cdot \frac{(n!)^2}{(2n)!}. \]
Cancelling the common factors:
\[ \frac{a_{n+1}}{a_n} = \frac{(2n+2)(2n+1)}{(n+1)^2}. \]
Since \(2n+2=2(n+1)\), it follows that
\[ \frac{a_{n+1}}{a_n} = \frac{2(n+1)(2n+1)}{(n+1)^2} = 2\frac{2n+1}{n+1}. \]
We evaluate the limit:
\[ \lim_{n\to+\infty}2\frac{2n+1}{n+1} = 2\lim_{n\to+\infty} \frac{2+\displaystyle\frac{1}{n}}{1+\displaystyle\frac{1}{n}} = 4. \]
Hence
\[ \lim_{n\to+\infty}\frac{a_{n+1}}{a_n}=4. \]
Since \(a_n>0\) and the limit of the ratio of consecutive terms exists and equals \(4>0\), the Stolz–Cesàro corollary on the \(n\)-th root implies
\[ \lim_{n\to+\infty}\sqrt[n]{a_n}=4. \]
Therefore
\[ \lim_{n\to+\infty}\sqrt[n]{\frac{(2n)!}{(n!)^2}}=4. \]
Exercise 15 — level ★★★★☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{\sum_{k=1}^{n}\log\left(1+\frac{1}{k}\right)}{\log n}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\sum_{k=1}^{n}\log\left(1+\frac{1}{k}\right)}{\log n}=1. \]
Solution
Set
\[ a_n=\sum_{k=1}^{n}\log\left(1+\frac{1}{k}\right), \qquad b_n=\log n. \]
We consider the ratio for \(n\ge 2\). Then \(b_n=\log n>0\), \(b_n\) is strictly increasing, and
\[ \lim_{n\to+\infty}\log n=+\infty. \]
The hypotheses on the denominator are therefore satisfied.
We compute the increments. For the numerator:
\[ a_{n+1}-a_n= \log\left(1+\frac{1}{n+1}\right). \]
For the denominator:
\[ b_{n+1}-b_n=\log(n+1)-\log n = \log\left(\frac{n+1}{n}\right) = \log\left(1+\frac{1}{n}\right). \]
The ratio of increments is
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{\log\left(1+\displaystyle\frac{1}{n+1}\right)} {\log\left(1+\displaystyle\frac{1}{n}\right)}. \]
To evaluate this limit, we multiply and divide so as to use the fundamental limit of the logarithm:
\[ \frac{\log\left(1+\displaystyle\frac{1}{n+1}\right)} {\log\left(1+\displaystyle\frac{1}{n}\right)} = \frac{ \displaystyle\frac{\log\left(1+\displaystyle\frac{1}{n+1}\right)}{\displaystyle\frac{1}{n+1}} }{ \displaystyle\frac{\log\left(1+\displaystyle\frac{1}{n}\right)}{\displaystyle\frac{1}{n}} } \cdot \frac{n}{n+1}. \]
Since
\[ \frac{\log(1+x)}{x}\to 1 \qquad \text{as } x\to 0, \]
we obtain
\[ \frac{ \displaystyle\frac{\log\left(1+\displaystyle\frac{1}{n+1}\right)}{\displaystyle\frac{1}{n+1}} }{ \displaystyle\frac{\log\left(1+\displaystyle\frac{1}{n}\right)}{\displaystyle\frac{1}{n}} } \to 1. \]
Moreover,
\[ \frac{n}{n+1}\to 1. \]
Hence
\[ \lim_{n\to+\infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = 1. \]
By the Stolz–Cesàro theorem it follows that
\[ \lim_{n\to+\infty}\frac{\sum_{k=1}^{n}\log\left(1+\frac{1}{k}\right)}{\log n} = 1. \]
We observe that in this problem the correct denominator is not \(n\), but \(\log n\). This choice is essential in order to obtain a finite, non-zero limit.
Exercise 16 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{\log n}{n}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\log n}{n}=0. \]
Solution
Set
\[ a_n=\log n, \qquad b_n=n. \]
We consider the ratio for \(n\ge 1\). The sequence \(b_n=n\) is positive for \(n\ge 1\), strictly increasing, and
\[ \lim_{n\to+\infty}b_n=+\infty. \]
We may therefore apply the Stolz–Cesàro theorem.
We compute the increments. For the numerator:
\[ a_{n+1}-a_n=\log(n+1)-\log n. \]
Using the properties of logarithms, we obtain
\[ a_{n+1}-a_n=\log\left(\frac{n+1}{n}\right)=\log\left(1+\frac{1}{n}\right). \]
For the denominator:
\[ b_{n+1}-b_n=(n+1)-n=1. \]
Hence
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \log\left(1+\frac{1}{n}\right). \]
Since
\[ \lim_{n\to+\infty}\log\left(1+\frac{1}{n}\right)=\log 1=0, \]
by the Stolz–Cesàro theorem it follows that
\[ \lim_{n\to+\infty}\frac{\log n}{n}=0. \]
This result expresses the fact that the logarithm grows more slowly than \(n\).
Exercise 17 — level ★★★☆☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{\sum_{k=1}^{n}\sqrt{k}}{\sum_{k=1}^{n}k}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\sum_{k=1}^{n}\sqrt{k}}{\sum_{k=1}^{n}k}=0. \]
Solution
Set
\[ a_n=\sum_{k=1}^{n}\sqrt{k}, \qquad b_n=\sum_{k=1}^{n}k. \]
The sequence \(b_n\) is positive for \(n\ge 1\). Moreover, it is strictly increasing, since
\[ b_{n+1}-b_n=n+1>0, \]
and it diverges to \(+\infty\), being the sum of the first \(n\) positive integers.
We may therefore apply the Stolz–Cesàro theorem.
We compute the increments:
\[ a_{n+1}-a_n=\sqrt{n+1}, \qquad b_{n+1}-b_n=n+1. \]
The ratio of increments is
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{\sqrt{n+1}}{n+1}. \]
Simplifying, we obtain
\[ \frac{\sqrt{n+1}}{n+1} = \frac{1}{\sqrt{n+1}}. \]
Since
\[ \lim_{n\to+\infty}\frac{1}{\sqrt{n+1}}=0, \]
the Stolz–Cesàro theorem implies
\[ \lim_{n\to+\infty}\frac{a_n}{b_n}=0. \]
Therefore
\[ \lim_{n\to+\infty}\frac{\sum_{k=1}^{n}\sqrt{k}}{\sum_{k=1}^{n}k}=0. \]
The result accords with the idea that the sum of the integers grows more rapidly than the sum of their square roots.
Exercise 18 — level ★★★★☆
Evaluate the limit
\[ \lim_{n\to+\infty}\frac{1^{3/2}+2^{3/2}+\cdots+n^{3/2}}{n^{5/2}}. \]
Answer
\[ \lim_{n\to+\infty}\frac{1^{3/2}+2^{3/2}+\cdots+n^{3/2}}{n^{5/2}}=\frac{2}{5}. \]
Solution
Set
\[ a_n=1^{3/2}+2^{3/2}+\cdots+n^{3/2}, \qquad b_n=n^{5/2}. \]
For \(n\ge 1\) we have \(b_n>0\). Moreover, \(b_n=n^{5/2}\) is strictly increasing and
\[ \lim_{n\to+\infty}b_n=+\infty. \]
The hypotheses on the denominator required by the Stolz–Cesàro theorem are therefore satisfied.
We compute the increments. For the numerator:
\[ a_{n+1}-a_n=(n+1)^{3/2}. \]
For the denominator:
\[ b_{n+1}-b_n=(n+1)^{5/2}-n^{5/2}. \]
The ratio of increments is
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{(n+1)^{3/2}}{(n+1)^{5/2}-n^{5/2}}. \]
Dividing numerator and denominator by \((n+1)^{3/2}\):
\[ \frac{(n+1)^{3/2}}{(n+1)^{5/2}-n^{5/2}} = \frac{1}{(n+1)-n\left(\displaystyle\frac{n}{n+1}\right)^{3/2}}. \]
We examine the denominator
\[ (n+1)-n\left(\frac{n}{n+1}\right)^{3/2}. \]
Set
\[ t_n=\frac{1}{n+1}. \]
Then \(t_n\to 0^+\) and
\[ \frac{n}{n+1}=1-t_n. \]
Moreover, \(n+1=\frac{1}{t_n}\) and \(n=\frac{1-t_n}{t_n}\). Hence
\[ (n+1)-n\left(\frac{n}{n+1}\right)^{3/2} = \frac{1}{t_n}-\frac{1-t_n}{t_n}(1-t_n)^{3/2}. \]
Hence
\[ (n+1)-n\left(\frac{n}{n+1}\right)^{3/2} = \frac{1-(1-t_n)^{5/2}}{t_n}. \]
Using the fundamental limit, equivalent to the differentiability of the function \(x^{5/2}\) at \(x=1\),
\[ \lim_{t\to 0^+}\frac{1-(1-t)^{5/2}}{t}=\frac{5}{2}. \]
It follows that
\[ (n+1)-n\left(\frac{n}{n+1}\right)^{3/2}\to\frac{5}{2}. \]
Hence
\[ \lim_{n\to+\infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{1}{\frac{5}{2}} = \frac{2}{5}. \]
By the Stolz–Cesàro theorem we conclude that
\[ \lim_{n\to+\infty}\frac{1^{3/2}+2^{3/2}+\cdots+n^{3/2}}{n^{5/2}} = \frac{2}{5}. \]
Exercise 19 — level ★★★★☆
Evaluate the limit
\[ \lim_{n\to+\infty}\sqrt[n]{\frac{(3n)!}{(n!)^3}}. \]
Answer
\[ \lim_{n\to+\infty}\sqrt[n]{\frac{(3n)!}{(n!)^3}}=27. \]
Solution
Set
\[ a_n=\frac{(3n)!}{(n!)^3}. \]
For every \(n\ge 1\) we have \(a_n>0\). We may therefore apply the corollary concerning the \(n\)-th root, after examining the ratio of consecutive terms.
We compute
\[ \frac{a_{n+1}}{a_n} = \frac{\displaystyle\frac{(3n+3)!}{((n+1)!)^3}} {\displaystyle\frac{(3n)!}{(n!)^3}}. \]
Dividing by a fraction amounts to multiplying by its reciprocal:
\[ \frac{a_{n+1}}{a_n} = \frac{(3n+3)!}{((n+1)!)^3} \cdot \frac{(n!)^3}{(3n)!}. \]
We now use the identities
\[ (3n+3)!=(3n+3)(3n+2)(3n+1)(3n)! \]
and
\[ (n+1)!=(n+1)n!. \]
From the latter it follows that
\[ ((n+1)!)^3=(n+1)^3(n!)^3. \]
Substituting, we obtain
\[ \frac{a_{n+1}}{a_n} = \frac{(3n+3)(3n+2)(3n+1)(3n)!}{(n+1)^3(n!)^3} \cdot \frac{(n!)^3}{(3n)!}. \]
Cancelling the common factors:
\[ \frac{a_{n+1}}{a_n} = \frac{(3n+3)(3n+2)(3n+1)}{(n+1)^3}. \]
We evaluate the limit by dividing each factor by \(n\):
\[ \frac{(3n+3)(3n+2)(3n+1)}{(n+1)^3} = \frac{ \left(3+\displaystyle\frac{3}{n}\right) \left(3+\displaystyle\frac{2}{n}\right) \left(3+\displaystyle\frac{1}{n}\right)} {\left(1+\displaystyle\frac{1}{n}\right)^3}. \]
Hence
\[ \lim_{n\to+\infty}\frac{a_{n+1}}{a_n} = \frac{3\cdot 3\cdot 3}{1^3} = 27. \]
Since \(a_n>0\) and the limit of the ratio of consecutive terms exists and equals \(27>0\), the Stolz–Cesàro corollary on the \(n\)-th root implies
\[ \lim_{n\to+\infty}\sqrt[n]{a_n}=27. \]
Substituting the definition of \(a_n\), we obtain
\[ \lim_{n\to+\infty}\sqrt[n]{\frac{(3n)!}{(n!)^3}}=27. \]
Exercise 20 — level ★★★★★
Evaluate the limit
\[ \lim_{n\to+\infty} \frac{\sum_{k=1}^{n}\log k}{\sum_{k=1}^{n}\log(k^2+k)}. \]
Answer
\[ \lim_{n\to+\infty} \frac{\sum_{k=1}^{n}\log k}{\sum_{k=1}^{n}\log(k^2+k)} = \frac{1}{2}. \]
Solution
Set
\[ a_n=\sum_{k=1}^{n}\log k, \qquad b_n=\sum_{k=1}^{n}\log(k^2+k). \]
For every \(k\ge 1\) we have \(k^2+k\ge 2\), hence
\[ \log(k^2+k)\ge \log 2>0. \]
Consequently \(b_n>0\) for every \(n\ge 1\). Moreover, \(b_n\) is strictly increasing, since
\[ b_{n+1}-b_n=\log((n+1)^2+(n+1))>0. \]
Finally,
\[ b_n=\sum_{k=1}^{n}\log(k^2+k)\ge n\log 2\to+\infty. \]
We may therefore apply the Stolz–Cesàro theorem.
We compute the increments. For the numerator:
\[ a_{n+1}-a_n=\log(n+1). \]
For the denominator:
\[ b_{n+1}-b_n=\log((n+1)^2+(n+1)). \]
Since
\[ (n+1)^2+(n+1)=(n+1)(n+2), \]
we obtain
\[ b_{n+1}-b_n=\log((n+1)(n+2)). \]
Using the property of the logarithm of a product:
\[ b_{n+1}-b_n=\log(n+1)+\log(n+2). \]
Hence the ratio of increments is
\[ \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{\log(n+1)}{\log(n+1)+\log(n+2)}. \]
Dividing numerator and denominator by \(\log(n+1)\):
\[ \frac{\log(n+1)}{\log(n+1)+\log(n+2)} = \frac{1}{1+\displaystyle\frac{\log(n+2)}{\log(n+1)}}. \]
Now
\[ \frac{\log(n+2)}{\log(n+1)} = \frac{\log(n+1)+\log\left(1+\displaystyle\frac{1}{n+1}\right)}{\log(n+1)}. \]
Hence
\[ \frac{\log(n+2)}{\log(n+1)} = 1+ \frac{\log\left(1+\displaystyle\frac{1}{n+1}\right)}{\log(n+1)}. \]
Since
\[ \log\left(1+\frac{1}{n+1}\right)\to 0 \]
and
\[ \log(n+1)\to+\infty, \]
it follows that
\[ \frac{\log\left(1+\displaystyle\frac{1}{n+1}\right)}{\log(n+1)}\to 0. \]
Hence
\[ \frac{\log(n+2)}{\log(n+1)}\to 1. \]
Hence
\[ \lim_{n\to+\infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{1}{1+1} = \frac{1}{2}. \]
By the Stolz–Cesàro theorem we conclude that
\[ \lim_{n\to+\infty} \frac{\sum_{k=1}^{n}\log k}{\sum_{k=1}^{n}\log(k^2+k)} = \frac{1}{2}. \]