<s>[INST] Problem with proof by induction I am struggling with the following equation, which I need to proof by induction:
$$\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}= \sum_{k=n+1}^{2n}\frac{1}{k}$$
$n\in \mathbb{N}$. I tried a few times and always got stuck.
Help would be appreciated. 
 [/INST] The proof is simple. By induction hypothesis, assume that the statement is true for $s < n$. Then, we have
\begin{align}
\sum\limits_{k = 1}^{2n} \dfrac{\left( 1 \right)^{k + 1}}{k} &= \sum\limits_{k = 1}^{2 \left( n - 1 \right)} \dfrac{\left( 1 \right)^{k + 1}}{k} + \dfrac{1}{2n - 1} - \dfrac{1}{2n} \\
&= \sum\limits_{k = n}^{2 \left( n - 1 \right)} \dfrac{1}{k} + \dfrac{1}{2n - 1} - \dfrac{1}{2n} \\
&= \sum\limits_{k = n + 1}^{2 \left( n - 1 \right)} \dfrac{1}{k} + \dfrac{1}{2n - 1} + \dfrac{1}{n} - \dfrac{1}{2n} \\
&= \sum\limits_{k = n + 1}^{2n} \dfrac{1}{k}
\end{align}
which completes the proof!
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