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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:

  1. How do you get from step $(1)$ to step $(2)$?
  2. How do you get from $(2)$ to $(3)$
  3. Is there a PDF that lists all the summation rules and identities?
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When you face problem like this, always try to decompose the summation into simple addition form. If you decompose your summation into simple series addition form these summations are settled down easily.

Step 1-2: \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 1{2k-1}+\frac 12\sum\limits_{k=0}^{n-1}\frac 1{2k+1}-\frac12+\frac12\cdot\frac{1}{2n+1}\\ &=\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}-\frac12\left(1-\frac{1}{2n+1}\right)\\ &=\sum\limits_{k=1}^n\frac 1{2k-1}-\frac{n}{2n+1}\\ \implies&\boxed{\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}-\frac12\sum\limits_{k=1}^n\frac1k+\frac{n}{2n+1}=\sum\limits_{k=1}^n\frac 1{2k-1}-\frac12\sum\limits_{k=1}^n\frac1k} \end{align*} Step 2-3: \begin{align*} &\sum\limits_{k=1}^n\frac 1{2k-1}\\ &=\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k}-\sum\limits_{k=1}^n\frac 1{2k}\\ &=\sum\limits_{k=1}^{2n}\frac1k-\frac12\sum\limits_{k=1}^n\frac 1k\\ \implies&\boxed{\sum\limits_{k=1}^n\frac 1{2k-1}-\frac12\sum\limits_{k=1}^n\frac 1k=\sum\limits_{k=1}^{2n}\frac1k-\sum\limits_{k=1}^n\frac1k=\sum\limits_{k=1}^n\frac 1{k+n}} \end{align*}

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  • $\begingroup$ Is there a PDF where I can review over all the rules and identities for summation? Such as$$\sum\limits_{k=1}^{n}\text{something}\rightarrow\sum\limits_{k=0}^{n-1}\text{something}$$For some reason, the manipulations aren't "clicking" into place $\endgroup$ – Crescendo May 25 '17 at 21:53
  • $\begingroup$ I don't think so; there is no particular rule, you've to do it manually, try like this: $\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}\\ =\frac12\left(1+\frac13+\frac15+\cdots+\frac{1}{2n-1}\right)+\frac12\left(\frac13+\frac15+\cdots+\frac{1}{2n-1}+\frac{1}{2n+1}\right)\\ =\frac12\left(1+\frac13+\frac15+\cdots+\frac{1}{2n-1}\right)+\frac12\left(1+\frac13+\frac15+\cdots+\frac{1}{2n-1}\right)-\frac12\cdot1+\frac12\cdot\frac{1}{2n+1}\\ =\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=0}^{n-1}\frac 1{2k+1}-\frac{n}{2n+1}$ $\endgroup$ – k.Vijay May 26 '17 at 3:37
  • $\begingroup$ You can also follow this: Rules for Product and Summation notation $\endgroup$ – k.Vijay May 26 '17 at 3:44
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  1. 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*}

  1. 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*}

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\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 $(3)$

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All in one chain: $$\color{blue}{\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}-\frac 12\sum\limits_{k=1}^n\frac 1k+\frac n{2n+1}=}$$ $$\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\left(\sum\limits_{k=1}^n\frac 1{2k-1}-1+\frac{1}{2n+1}\right)-\frac 12\sum\limits_{k=1}^n\frac 1k+\frac n{2n+1}=$$ $$\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\left(-\frac{2n}{2n+1}\right)-\frac 12\sum\limits_{k=1}^n\frac 1k+\frac n{2n+1}=$$

$$\color{blue}{\sum\limits_{k=1}^n\frac 1{2k-1}-\frac 12\sum\limits_{k=1}^n\frac 1k=}$$ $$\left(\sum\limits_{k=1}^{2n}\frac 1{k}-\sum\limits_{k=1}^n\frac 1{2k}\right)-\frac 12\sum\limits_{k=1}^n\frac 1k=$$ $$\left(\sum\limits_{k=1}^{2n}\frac 1{k}-\frac12\sum\limits_{k=1}^n\frac 1{k}\right)-\frac 12\sum\limits_{k=1}^n\frac 1k=$$

$$\color{blue}{\sum\limits_{k=1}^{2n}\frac 1k-\sum\limits_{k=1}^n\frac 1k=}$$ $$\left(\sum\limits_{k=1}^{n}\frac 1k+\sum\limits_{k=n+1}^{2n}\frac 1k\right)-\sum\limits_{k=1}^n\frac 1k=$$ $$\sum\limits_{k=n+1}^{2n}\frac 1k=$$

$$\color{blue}{\sum\limits_{k=1}^n\frac 1{k+n}}$$

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