Manipulating a Summation Series 
Summation: How does$$\begin{align*} & \frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}-\frac 12\sum\limits_{k=1}^n\frac 1k+\frac n{2n+1}\tag1\\ & =\sum\limits_{k=1}^n\frac 1{2k-1}-\frac 12\sum\limits_{k=1}^n\frac 1k\tag2\\ & =\sum\limits_{k=1}^{2n}\frac 1k-\sum\limits_{k=1}^n\frac 1k=\sum\limits_{k=1}^n\frac 1{k+n}\tag3\end{align*}$$
  Given that$$\sum\limits_{k=1}^n\frac 1{(2k)^3-2k}=\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}-\frac 12\sum\limits_{k=1}^n\frac 1k$$

I'm not sure how they got from the first step to the second. The $\tfrac 12\sum\limits_{k=1}^n\tfrac 1k$ didn't change, so that must mean that$$\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}+\frac n{2n+1}=\sum\limits_{k=1}^n\frac 1{2k-1}$$But I don't see how. I tried expanding the LHS, and rearranging to get the RHS, but that didn't go far.$$\begin{align*}\frac 12\left(\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k+1}\right)+\frac n{2n+1} & =\frac 12\left(\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 1{2k+1}\right)+\frac n{2n+1}\\ & =\frac 12\sum\limits_{k=1}^n\frac {4k}{4k^2-1}+\frac n{2n+1}\\ & =2\sum\limits_{k=1}^n\frac k{(2k)^2-1}+\frac n{2n+1}\end{align*}$$I feel like I'm close, but I just can't seem to finish it. I believe that you'll need to use a summation identity.

Questions:



*

*How do you get from step $(1)$ to step $(2)$?

*How do you get from $(2)$ to $(3)$

*Is there a PDF that lists all the summation rules and identities?


\begin{align}
&\sum\limits_{k=1}^n\frac 1{2k+1}\\
= & \   \sum\limits_{k=1}^{n-1}\frac 1{2k+1} + \frac{1}{2n+1} \\
= & \ 1 + \sum\limits_{k=1}^{n-1}\frac 1{2k+1} + \frac{1}{2n+1} -1\\
= & \ \sum\limits_{k=0}^{n-1}\frac 1{2k+1} + \frac{1 -2n-1}{2n+1}\\
= & \ \sum\limits_{l=1}^{n}\frac 1{2l-1} - \frac{2n}{n+1} \tag{where $l = k+1$}\\
= & \  \sum\limits_{k=1}^{n}\frac 1{2k-1} - \frac{2n}{n+1} \tag{$\ast$}
\end{align}
Therefore,
\begin{align}
 &\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}-\frac 12\sum\limits_{k=1}^n\frac 1k+\frac n{2n+1}\\ 
= & \ \frac 12\sum\limits_{k=1}^n\frac 1{2k-1} + \frac{1}{2}\left( \sum\limits_{k=1}^{n}\frac 1{2k-1} - \frac{2n}{2n+1} \right) -\frac 12\sum\limits_{k=1}^n\frac 1k + \frac n{2n+1}\tag{using $(\ast)$}\\
= & \ \sum\limits_{k=1}^n\frac 1{2k-1} -\frac 12\sum\limits_{k=1}^n\frac 1k
\end{align}
This gives $(1)$ to $(2)$.
Now,
\begin{align}
&\sum_{k=1}^{n}\frac{1}{2k-1} \\
= & \ \sum_{k=1}^n\frac{1}{2k-1} + \sum_{k=1}^n\frac{1}{2k} - \sum_{k=1}^n\frac{1}{2k}\\
= & \ \sum_{k=1}^{2n} \frac{1}{k}- \frac{1}{2}\sum_{k=1}^n\frac{1}{k}\\
\end{align}
And then,
\begin{align}
&\sum\limits_{k=1}^n\frac 1{2k-1} -\frac 12\sum\limits_{k=1}^n\frac 1k \\
= & \ \sum_{k=1}^{2n} \frac{1}{k}- \frac{1}{2}\sum_{k=1}^n\frac{1}{k} -\frac 12\sum\limits_{k=1}^n\frac 1k\\
= & \ \sum_{k=1}^{2n} \frac{1}{k}-  \sum_{k=1}^n\frac{1}{k} \\
= & \ \sum_{k=n+1}^{2n}\frac{1}{k} = \sum_{k=1}^{n}\frac{1}{n+k}
\end{align}
This gives $(2)$ to $(3)$
