\(\sum_{k=n+1}^{mn}\frac{1}{k}\)

[Grigorios Kostakos]

For fixed \(m\in\mathbb{N}\,,\;m\geqslant2\,,\) find the limit of the sequence \[a_{n}=\displaystyle\mathop{\sum}\limits_{k=n+1}^{mn}\frac{1}{k}\,,\quad n\in\mathbb{N}\,.\]

[admin]

Replied by ex-member aziiri:

Using MVT for the \(\ln\) function we get : \[ \ln(k+1)-\ln k\leq\frac{1}{k}\leq\ln k-\ln(k-1) .\] Now we do the sum from \(k=n+1\) to \(m n\) (telescopic sum), to get : $$\ln(mn+1)-\ln (n+1) \leq \sum_{k=n+1}^{m n} \frac{1}{k} \leq \ln m n - \ln n =\ln m.$$
It is obvious that the left side tends towards \(\ln m\), so the requested limit is \(\ln m\) by the squeeze theorem.

[Grigorios Kostakos]

Another solution:

\begin{align*}
a_{n}&=\displaystyle\mathop{\sum}\limits_{k=n+1}^{mn}\frac{1}{k}=\mathop{\sum}\limits_{k=1}^{mn}\frac{1}{k}-\mathop{\sum}\limits_{k=1}^{n}\frac{1}{k}\\
&=\mathop{\sum}\limits_{k=1}^{mn}\frac{1}{k}-\log(mn)-\biggl({\mathop{\sum}\limits_{k=1}^{n}\frac{1}{k}-\log{n}}\biggr)+\log{m}\,,\quad n\in\mathbb{N}\,,\;m\in\mathbb{N}\,,\;m\geqslant2\,.
\end{align*} But \[\displaystyle\mathop{\lim}\limits_{n\rightarrow{+\infty}}\biggl({\mathop{\sum}\limits_{k=1}^{mn}{\frac{1}{k}}-\log(mn)}\biggr)\,\mathop{=\!=\!=\!=}\limits^{{\begin{subarray}{c}
{r\,=\,mn} \\
{r\to0} \\
\end{subarray}}}\,\mathop{\lim}\limits_{r\rightarrow{+\infty}}\biggl({\mathop{\sum}\limits_{k=1}^{r}{\frac{1}{k}}-\log{r}}\biggr)=\gamma\,,\] where \(\gamma\) is the Euler-Mascheroni constant.

So \begin{align*}
\displaystyle\mathop{\lim}\limits_{n\rightarrow{+\infty}}{a_n}&=\mathop{\lim}\limits_{n\rightarrow{+\infty}}\biggl({\mathop{\sum}\limits_{k=1}^{mn}\frac{1}{k}-\log(mn)}\biggr)-\mathop{\lim}\limits_{n\rightarrow{+\infty}}\biggl({\mathop{\sum}\limits_{k=1}^{n}\frac{1}{k}-\log{n}}\biggr)+\mathop{\lim}\limits_{n\rightarrow{+\infty}}\log{m}\\
&=\gamma-\gamma+\log{m}=\log{m}\,.
\end{align*}