Summation Notation using Induction $$\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$$
I am trying to prove this inductively, so I thought that I would expand the right side out of sigma form to get
$$\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k} = \frac{2}{2n(2n+1)} - \frac{1}{n}$$
which simplified to
$$\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k} = \frac{-2}{2n+1}$$
but apparently that isn't correct, can someone provide some insight into what I am doing wrong?
 A: The right side doesn't equal what you think it equals.  I think maybe you're confusing this with the different, and unrelated, formula $\sum_{k=1}^n k = \frac{n(n+1)}{2}.$  You can't get from that formula to a formula for $\sum_{k=1}^n \frac1{k}$ or anything similar.
See https://math.stackexchange.com/a/1961771/35021 for a detailed answer to your question.  (The question in the other post is phrased as a limit, but the answer there solves your problem exactly.)
A: 
INDUCTION PROOF:

If one wishes to prove the identity using induction, we first establish a benchmark.  Letting $n=1$, we find that $\sum_{k=1}^{2}\frac{(-1)^{k+1}}{k}=1/2=\sum_{k=2}^2\frac1k$.
Next, we assume that the identity works for some number $n$
$$\color{blue}{\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}}=\color{red}{\sum_{k=n+1}^{2n}\frac1k} \tag 1$$
Then, we test the validity of the relationship in $(1)$ for $n+1$.  Proceeding we find that
$$\begin{align}
\sum_{k=1}^{2n+2}\frac{(-1)^{k+1}}{k}&=\color{blue}{\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}}+\frac{1}{2n+1}-\frac{1}{2n+2}\\\\
&=\color{red}{\sum_{k=n+1}^{2n}\frac1k}+\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)\\\\
&=\sum_{k=n+2}^{2n+2}\frac1k+\frac{1}{n+1}-\left(\frac{1}{2n+1}+\frac{1}{2n+2}\right)+\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)\\\\
&=\sum_{k=n+2}^{2n+2}\frac1k
\end{align}$$
as expected.


DIRECT PROOF:

Note that we can write
$$\begin{align}
\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}&=\sum_{k=1}^n\frac{1}{2k-1}-\sum_{k=1}^n\frac1{2k}\\\\
&=\sum_{k=1}^n\left(\frac{1}{2k-1}+\frac{1}{2n}\right)-2\sum_{k=1}^{n}\frac1{2k}\\\\
&=\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1{k}\\\\
&=\sum_{k=n+1}^{2n}\frac1k
\end{align}$$
as was to be shown!
