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?


*

*Note that


$$\sum_{k=1}^n\frac{1}{2k+1}=\sum_{k=2}^{n+1}\frac{1}{2k-1}=\sum_{k=1}^{n}\frac{1}{2k-1}-1+\frac{1}{2n+1}$$
Therefore,
\begin{align*}
\frac{1}{2}\sum_{k=1}^n\frac{1}{2k-1}+\frac{1}{2}\sum_{k=1}^n\frac{1}{2k+1}+\frac{n}{2n+1}&=\sum_{k=1}^n\frac{1}{2k-1}-\frac{1}{2}+\frac{1}{2(2n+1)}+\frac{n}{2n+1}\\
&=\sum_{k=1}^n\frac{1}{2k-1}
\end{align*}


*Note that


$$\sum_{k=1}^n\frac{1}{2k-1}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{2k}=\sum_{k=1}^{2n}\frac{1}{k}-\frac{1}{2}\sum_{k=1}^n\frac{1}{k}$$
Therefore,
\begin{align*}
\sum_{k=1}^n\frac{1}{2k-1}-\frac{1}{2}\sum_{k=1}^n\frac{1}{k}&=\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}{k+n}\\
\end{align*}
