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. 
 A: Base case of condition is quite clearly satisfied. Now assume that the result holds for some n. We have to show that the result holds for n+1. 
$$\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}= \sum_{k=n+1}^{2n}\frac{1}{k}$$ 
The above expression is our assumption and we have to show,
$$\sum_{k=1}^{2n+2}\frac{(-1)^{k+1}}{k}= \sum_{k=n+2}^{2n+2}\frac{1}{k}$$
So, consider the LHS of the above equation
$$\sum_{k=1}^{2n+2}\frac{(-1)^{k+1}}{k}= \sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}+\frac{1}{2n+1}+\frac{-1}{2n+2}$$
$$\sum_{k=1}^{2n+2}\frac{(-1)^{k+1}}{k}= \sum_{k=n+1}^{2n}\frac{1}{k}+\frac{1}{2n+1}+\frac{-1}{2n+2}$$
$$\sum_{k=1}^{2n+2}\frac{(-1)^{k+1}}{k}= \sum_{k=n+2}^{2n}\frac{1}{k}+\frac{1}{n+1}+\frac{1}{2n+1}+\frac{-1}{2n+2}$$
$$\sum_{k=1}^{2n+2}\frac{(-1)^{k+1}}{k}= \sum_{k=n+2}^{2n+2}\frac{1}{k}$$
Hence proved.
A: 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!
