show that $\sum_{i = 1}^{2k}\frac{ (-1)^{i+1}}{i} = \sum_{i = k+1}^{2k} \frac{1}{i}$ I have a proof but it does not seem elegant. Is there a more elegant solution? Thanks.
Consider $X = \sum_{i = 1}^{2k}\frac{(-1)^{i+1}}{i} = X_1 + X_2$ where
$X_1 = 1 - \frac{1}{2} + ... + \frac{1}{k-1} - \frac{1}{k}$
$X_2 = \frac{1}{k+1} - \frac{1}{k+2} + ... + \frac{1}{2k-1} - \frac{1}{2k}$
$k$ is even. 
Every negative term $d$ in $X_2$ is of the form $\frac{1}{2^my}$ where $y$ is odd and could be $1$. Also $y \le k$ 
For every such $d$, we will have the following terms 
$\frac{1}{y} - \frac{1}{2y} - ... - \frac{1}{2^{m-1}y}$ in $X_1$. 
Adding $d$ to the above, we get
$\frac{1}{y} - \frac{1}{2y} - ... - \frac{1}{2^{m-1}y} - \frac{1}{2^my} = \frac{1}{2^my}$
So every negative term $d$ in $X_2$ is now replaced with a term of the same magnitude but with a positive sign. 
Hence $X = \frac{1}{k+1} + \frac{1}{k+2} + ... + \frac{1}{2k-1} + \frac{1}{2k}$

We can use harmonic numbers 
$H_k=\sum_{i=1}^k\frac{1}{i}$.

We obtain
  \begin{align*}
\sum_{i=k+1}^{2k}\frac{1}{i}=\color{blue}{H_{2k}-H_k}
\end{align*}
on the other hand we get
  \begin{align*}
\sum_{i=1}^{2k}\frac{(-1)^{i+1}}{i}-H_{2k}
&=-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2k}\right)=-H_k\\
\end{align*}
which implies
  \begin{align*}
\sum_{i=1}^{2k}\frac{(-1)^{i+1}}{i}
&=\color{blue}{H_{2k}-H_k}
\end{align*}
  and the claim follows.

