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 $X = \frac{1}{k+1} + \frac{1}{k+2} + ... + \frac{1}{2k-1} + \frac{1}{2k}$
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\sum_{i = 1}^{2k}{\pars{-1}^{i + 1} \over i} =
\sum_{i = k + 1}^{2k}{1 \over i}:\ {\large ?}}$.

\begin{align}
\sum_{i = 1}^{2k}{\pars{-1}^{i + 1} \over i} & =
\sum_{i = 1}^{k}{1 \over 2i - 1} - \sum_{i = 1}^{k}{1 \over 2i} =
\pars{\sum_{i = 1}^{2k}{1 \over i} - \sum_{i = 1}^{k}{1 \over 2i}} -
\sum_{i = 1}^{k}{1 \over 2i}
\\[5mm] & =
\sum_{i = 1}^{2k}{1 \over i} - \sum_{i = 1}^{k}{1 \over i} =
\bbx{\ds{\sum_{i = k + 1}^{2k}{1 \over i}}}
\end{align}
A: Let $S_k=\sum_{k=1}^{2k}(-1)^{i+1}/i$ and $T_k=\sum_{i+1}^{2i}1/i$. Then
$$T_k-T_{k-1}=\frac1{2k-1}+\frac1{2k}-\frac1k=\frac1{2k-1}-\frac1{2k}=
S_k-S_{k-1}.$$
Obviously, $S_0=T_0$ so by induction....
A: You're right, there's a much simpler way:
$$S=\sum^{2k}_{i=1}\left[\frac{(-1)^{(i+1)}}{i}\right]=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6} \cdot\cdot\cdot +\frac{1}{2k-1}-\frac{1}{2k}$$
$$S=\color{blue}{\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6} \cdot\cdot\cdot +\frac{1}{2k-1}+\frac{1}{2k}\right]}-2\color{red}{\left[\frac{1}{2}+\frac{1}{4}+\cdot\cdot\cdot\frac{1}{2k}\right]}$$
$$S=\sum^{2k}_{i=1}\frac{1}{i}-2\sum^k_{i=1}\frac{1}{2i}=\sum^{2k}_{i=1}\frac{1}{i}-\sum^k_{i=1}\frac{2}{2i}=\color{purple}{\sum^{2k}_{i=1}\frac{1}{i}-\sum^k_{i=1}\frac{1}{i}}=\sum^{2k}_{i=k+1}\frac{1}{i}$$
Hence proved.

Note: $$\sum^{2k}_{i=1}\frac{1}{i}=\sum^{k}_{i=1}\frac{1}{i}+\sum^{2k}_{i=k+1}\frac{1}{i}$$
A: 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.

