<s>[INST] 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 $X = \frac{1}{k+1} + \frac{1}{k+2} + ... + \frac{1}{2k-1} + \frac{1}{2k}$
 [/INST] 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: $$\sum^{2k}_{i=1}\frac{1}{i}=\sum^{k}_{i=1}\frac{1}{i}+\sum^{2k}_{i=k+1}\frac{1}{i}$$
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