# tough series involving digamma

I ran across a series that is rather challenging. For kicks I ran it through Maple and it gave me a conglomeration involving digamma. Mathematica gave a solution in terms of Lerch Transcendent, which was worse yet. Perhaps residues would be a better method?.

But, it is $$\sum_{k=1}^{\infty}\frac{(-1)^{k}(k+1)}{(2k+1)^{2}-a^{2}}.$$

The answer Maple spit out was:

$$\frac{a+1}{16a}\left[\psi\left(\frac{3}{4}-\frac{a}{4}\right)-\psi\left(\frac{-a}{4}+\frac{1}{4}\right)\right]+\frac{a-1}{16a}\left[\psi\left(\frac{3}{4}+\frac{a}{4}\right)-\psi\left(\frac{1}{4}+\frac{a}{4}\right)\right]+\frac{1}{a^{2}-1}.$$

Is it possible to actually get to something like this by using $\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{k+a}\right]=\gamma+\psi(a+1)?$

I tried, but to no avail. But, then again, maybe it is too cumbersome.

i.e. I tried expanding it into

$\frac{k+1}{(2k+1)^{2}-a^{2}}=\frac{-1}{4(a-2k-1)(2k+1)}-\frac{1}{4(a-2k-1)}+\frac{1}{4(a+2k+1)(2k+1)}+\frac{1}{4(a+2k+1)}$

then using $\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{k-\frac{1}{4}-\frac{a}{4}}\right]=\psi\left(\frac{3}{4}-\frac{a}{4}\right)$ and so on, but it did not appear to be anywhere close to the given series.

On another point, can it be done using residues?. By using $$\frac{\pi csc(\pi z)(z+1)}{(2z+1)^{2}-a^{2}}.$$

This gave me a residue at $\frac{a-1}{2} and \frac{-(a+1)}{2}$ of

$\frac{-\pi}{a-1}sec(a\pi/2)$ and $\frac{\pi}{a+1}sec(\pi a/2)$

Taking the negative sum of the residues, it is $\frac{2\pi}{(a-1)(a+1)}sec(a\pi/2)$

By subbing in k=0 into the series, it gives $\frac{-1}{a^{2}-1}$.

I try adding them up and finding the sum, but it does not appear to work out.

Any suggestions?. Perhaps there is another method I am not trying?. There probably is. Thanks a million.

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First, group the consecutive oscillating terms together:

$$\sum_{k=1}^{\infty}\frac{(-1)^{k}(k+1)}{(2k+1)^{2}-a^{2}}=\sum_{k=0}^\infty \left(\frac{2k+1}{(4k+1)^2-a^2}-\frac{2k+2}{(4k+3)^2-a^2}\right)-\frac{2(0)+1}{(2(0)+1)^2-a^2}$$

Next, invoke partial fraction decomposition and solve for coefficients:

$$\frac{2k+1}{(4k+1)^2-a^2} = \frac{a+1}{16a}\frac{1}{k+\frac{1-a}{4}}+\frac{a-1}{16a}\frac{1}{k+\frac{1+a}{4}},$$

and similarly

$$\frac{2k+2}{(4k+3)^2-a^2}=\frac{a+1}{16a}\frac{1}{k+\frac{3-a}{4}}+\frac{a-1}{16a}\frac{1}{k+\frac{3+a}{4}}.$$

Hence we are left with

$$\frac{a+1}{16a}\sum_{k=0}^\infty \left(\frac{1}{k+\frac{1-a}{4}}-\frac{1}{k+\frac{3-a}{4}}\right)+\frac{a-1}{16a}\sum_{k=0}^\infty\left(\frac{1}{k+\frac{1+a}{4}}-\frac{1}{k+\frac{3+a}{4}}\right)+\frac{1}{a^2-1}$$

$$=\begin{array}{c} \frac{a+1}{16a}\sum_{k=0}^\infty \left(\left(\frac{1}{k+1}-\frac{1}{k+\frac{3-a}{4}}\right)-\left(\frac{1}{k+1}-\frac{1}{k+\frac{1-a}{4}}\right)\right) \\ +\frac{a-1}{16a}\sum_{k=0}^\infty\left(\left(\frac{1}{k+1}-\frac{1}{k+\frac{3+a}{4}}\right)-\left(\frac{1}{k+1}-\frac{1}{k+\frac{1+a}{4}}\right)\right)+\frac{1}{a^2-1} \end{array}$$

$$=\frac{a+1}{16a}\left[\psi\left(\frac{3-a}{4}\right)-\psi\left(\frac{1-a}{4}\right)\right]+\frac{a-1}{16a}\left[\psi\left(\frac{3+a}{4}\right)-\psi\left(\frac{1+a}{4}\right)\right]+\frac{1}{a^{2}-1}.$$

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Awesome, anon. You're very sharp. I did not think about breaking it up into oscillating terms like that. Also, that partial fraction you came up with is clever. I would not have come up with that. I did manage to factor it and solve for $\frac{a-1}{2}$ and $\frac{-a-1}{2}$. You solved another series I posted sometime back involving digamma and a series with ln. You are a wizard with it :):) Thanks a bunch. – Cody Jun 4 '12 at 21:32
@Cody: Please remember to use the comments for these kinds of posts - answers are only for answering the question :) – Zev Chonoles Jun 4 '12 at 21:45
Sorry about that. I thought one 'answered' when accepting an answer. I will remember next time. – Cody Jun 4 '12 at 21:48
Our answers match numerically at several points and graphically. I am working on showing directly that they are the same :-) – robjohn Jun 5 '12 at 11:48
With $\frac1k$ in the summand, did you mean to sum from $0$? If you did, then I believe if you change $\frac1k$ to $\frac{1}{k+1}$, you get $$\psi(x)+\gamma=\sum_{k=0}^\infty\left(\frac{1}{k+1}-\frac{1}{k+x}\right)$$ then your equations work since the $\gamma$'s cancel. – robjohn Jun 6 '12 at 13:22

Note that $$\frac{1}{2k+1-a}+\frac{1}{2k+1+a}=\frac{4k+2}{(2k+1)^2-a^2}\tag{1}$$ and $$\frac1a\left(\frac{1}{2k+1-a}-\frac{1}{2k+1+a}\right)=\frac{2}{(2k+1)^2-a^2}\tag{2}$$ Adding $(1)$ and $(2)$ and dividing by $4$ yields \begin{align} \frac{k+1}{(2k+1)^2-a^2} &=\frac{1+a}{8a}\frac{1}{k+(1-a)/2}-\frac{1-a}{8a}\frac{1}{k+(1+a)/2}\\ &=\hphantom{+ }\frac{1-a}{8a}\left(\frac1k-\frac{1}{k+(1+a)/2}\right)\\ &\hphantom{= }-\frac{1+a}{8a}\left(\frac1k-\frac{1}{k+(1-a)/2}\right)\\ &\hphantom{= }+\frac{1}{4k}\tag{3} \end{align}

Now, using $$\psi(a+1)+\gamma=\sum_{k=1}^\infty\frac{1}{k}-\frac{1}{k+a}\tag{4}$$ we get $$\frac12\psi\left(\frac{a}{2}+1\right)+\frac\gamma2=\sum_{k=1}^\infty\frac{1}{2k}-\frac{1}{2k+a}\tag{5}$$ and subtracting twice $(5)$ from $(4)$ gives $$\psi(a+1)-\psi\left(\frac{a}{2}+1\right)=\sum_{k=1}^\infty(-1)^{k-1}\left(\frac{1}{k}-\frac{1}{k+a}\right)\tag{6}$$ Furthermore, $$\log(2)=\sum_{k=1}^\infty(-1)^{k-1}\frac1k\tag{7}$$

Using $(3)$, $(6)$, and $(7)$, we get \begin{align} \sum_{k=1}^\infty(-1)^k\frac{k+1}{(2k+1)^2-a^2} &=-\frac{1-a}{8a}\left(\psi\left(\frac{3+a}{2}\right)-\psi\left(\frac{5+a}{4}\right)\right)\\ &\hphantom{= }+\frac{1+a}{8a}\left(\psi\left(\frac{3-a}{2}\right)-\psi\left(\frac{5-a}{4}\right)\right)\\ &\hphantom{= }-\frac14\log(2)\tag{8} \end{align}

Equivalence of Forms:

Using $(4)$, $(5)$, and $(7)$, we get \begin{align} \sum_{k=1}^\infty\frac{1}{2k-1}-\frac{1}{2k-1+a} &=\color{green}{\sum_{k=1}^\infty\frac{1}{2k-1}-\frac{1}{2k}}+\color{red}{\sum_{k=1}^\infty\frac{1}{2k}-\frac{1}{2k-1+a}}\\ &=\color{green}{\log(2)}+\color{red}{\frac12\psi\left(\frac{a+1}{2}\right)+\frac\gamma2}\tag{9} \end{align} Adding $(5)$ to $(9)$ yields \begin{align} \psi(a+1)+\gamma &=\hphantom{+}\log(2)+\frac12\psi\left(\frac{a+1}{2}\right)+\frac\gamma2\\ &\hphantom{= }+\frac12\psi\left(\frac{a}{2}+1\right)+\frac\gamma2\tag{10} \end{align} Rearranging $(10)$ shows that $$\psi(a)=\log(2)+\frac12\psi\left(\frac{a}{2}\right)+\frac12\psi\left(\frac{a+1}{2}\right)\tag{11}$$ Applying $(11)$ gives $$\psi\left(\frac{3+a}{2}\right)=\log(2)+\frac12\psi\left(\frac{3+a}{4}\right)+\frac12\psi\left(\frac{5+a}{4}\right)\tag{12}$$ and $$\psi\left(\frac{3-a}{2}\right)=\log(2)+\frac12\psi\left(\frac{3-a}{4}\right)+\frac12\psi\left(\frac{5-a}{4}\right)\tag{13}$$ Plug $(12)$ and $(13)$ into $(8)$ \begin{align} \sum_{k=1}^\infty(-1)^k\frac{k+1}{(2k+1)^2-a^2} &=\hphantom{+}\frac{a-1}{16a}\left(\psi\left(\frac{3+a}{4}\right)-\psi\left(\frac{5+a}{4}\right)\right)\\ &\hphantom{= }+\frac{a+1}{16a}\left(\psi\left(\frac{3-a}{4}\right)-\psi\left(\frac{5-a}{4}\right)\right)\\ &=\hphantom{+}\frac{a-1}{16a}\left(\psi\left(\frac{3+a}{4}\right)-\psi\left(\frac{1+a}{4}\right)-\frac{4}{1+a}\right)\\ &\hphantom{= }+\frac{a+1}{16a}\left(\psi\left(\frac{3-a}{4}\right)-\psi\left(\frac{1-a}{4}\right)-\frac{4}{1-a}\right)\\ &=\hphantom{+}\color{red}{\frac{a-1}{16a}\left(\psi\left(\frac{3+a}{4}\right)-\psi\left(\frac{1+a}{4}\right)\right)}\\ &\hphantom{= }\color{red}{+\frac{a+1}{16a}\left(\psi\left(\frac{3-a}{4}\right)-\psi\left(\frac{1-a}{4}\right)\right)}\\ &\hphantom{= }\color{red}{+\frac{1}{a^2-1}}\tag{14} \end{align}

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Wow, Robjohn. Thanks a million. Sorry I was not back sooner. I had not realized anyone else had answered. I always appreciate your input. Thanks again for the detailed solution. – Cody Jun 7 '12 at 12:13