Formula for sum of the reciprocals of integers

Give a definition of $\sum_{k=m}^{m+n} a_k$ .
Prove the following is true for $n \geq 1$ :
$\sum_{k=n+1}^{2n} \frac{1}{k} = \sum_{m=1}^{2n} \frac{(-1)^{m+1}}{m}.$

We define
$\sum_{k=m}^{m+n} a_k = a_m + a_{m+1} + \cdots + a_{m+n}.$
Proof. The proof is by induction. For the case $n = 1$ we have, on the left,

$\sum_{k=n+1}^{2n} \frac{1}{k} = \sum_{k=2}^2 \frac{1}{k} = \frac{1}{2}.$

On the right,

$\sum_{m=1}^{2n} \frac{(-1)^{m+1}}{m} = \sum_{m=1}^2 \frac{(-1)^{m+1}}{m} = 1 - \frac{1}{2} = \frac{1}{2}.$

Thus, the formula holds for the case $n = 1$ . Assume then that it holds for some $n = j \in \mathbb{Z}_{\geq 1}$ . Then we have,

$\begin{align*} &&\sum_{k=j+1}^{2j} \frac{1}{k} &= \sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m} \\ \implies && \left(\sum_{k=j+1}^{2j} \frac{1}{k}\right) + \frac{1}{2j+1} + \frac{1}{2j+2} &= \left(\sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m}\right) + \frac{1}{2j+1} + \frac{1}{2j+2} \\ \implies && \sum_{k=j+1}^{2(j+1)} \frac{1}{k} &= \left(\sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m}\right) + \frac{1}{2j+1} + \frac{1}{2j+2} \\ \implies && \frac{1}{j+1} + \left(\sum_{k=j+2}^{2(j+1)} \frac{1}{k} \right)&= \left(\sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m}\right) + \frac{1}{2j+1} + \frac{1}{2j+2} \\ \implies && \sum_{k=j+2}^{2(j+1)} \frac{1}{k} &= \left( \sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m} \right) + \frac{1}{2j+1} + \frac{1}{2(j+1)} - \frac{1}{j+1} \\ \implies && \sum_{k=j+2}^{2(j+1)} \frac{1}{k} &= \left( \sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m} \right) + \frac{1}{2j+1} - \frac{1}{2(j+1)} \\ \implies && \sum_{k=j+2}^{2(j+1)} \frac{1}{k} &= \sum_{m=1}^{2(j+1)} \frac{(-1)^{m+1}}{m}. \end{align*}$

Thus, the formula holds for $j+1$ if it holds for $j$ ; hence, we have shown that it holds for all $n \in \mathbb{Z}_{\geq 1}. \qquad \blacksquare$