Problem by proving a statement by induction I want to prove the following statement by induction for $n\leq 1$:
\begin{align*}
\sum_{k=n}^{2n-1}\frac{1}{k}= \sum_{k=1}^{2n-1}\frac{(-1)^{k+1}}{k}
\end{align*}
For $n=1$ its easy to see:
\begin{align*}
\sum_{k=1}^{2-1}\frac{1}{k}= \sum_{k=1}^{1}\frac{1}{k}= 1 = \sum_{k=1}^{1}\frac{(-1)^{2}}{1}
\end{align*}
For $n\mapsto n+1$
\begin{align*}
\sum_{k=n+1}^{2(n+1)-1}\frac{1}{k}= \sum_{k=n+1}^{2n+1}\frac{1}{k} = \sum_{k=n-1}^{2n-1}\frac{1} {k}+\frac{1}{n+1}+\frac{1}{n}\dots \ ?
\end{align*}
I am not really sure how to deal with the $k=n+1$ in the sum, to use the case that the statement holds for n. Could anybody help me? Thank you very much!

Call the left sum $L(n)$ and the right sum $R(n)$. Then we have
$$L(n + 1) - L(n) = \frac{1}{2n}+\frac{1}{2n+1} - \frac{1}{n},$$
and
\begin{align}
R(n + 1) - R(n) &= \frac{(-1)^{2n}}{2n+1} + \frac{(-1)^{2n+2}}{2n+1} \\
&= - \frac{1}{2n} + \frac{1}{2n+1}.
\end{align}
So by comparison, we need to check that
$$\frac{1}{2n} - \frac{1}{n} = - \frac{1}{2n}, $$
which is easy enough.
Therefore $L(n+1) - L(n) = R(n + 1) - R(n)$. So if $L(n) = R(n)$ then $L(n + 1) = R(n + 1)$.
