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How to prove the equality below?

$$ \sum_{k=1}^{m} \frac{1}{(2k-1)^2}=\frac{\pi^2}{8}-\frac{1}{4}\psi^{(1)}\left(m+\frac{1}{2}\right)$$

When $\displaystyle m$ is a positive integer.

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    What is $\psi$? – 2017-01-29
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    @mathreadler The tri-gamma function. – 2017-01-29
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    Digamma function – 2017-01-29
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    @IsraelMeirelesChrisostomo Pretty sure this is wrong if it is the digamma function. – 2017-01-29
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    $ \psi$ is digamma function, $ \psi^{(1)}$ is tri-gamma function – 2017-01-29
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    [$\psi\,'(m+1/2) = \ ?$](https://en.wikipedia.org/wiki/Trigamma_function) and you need to know that $\frac{\pi^2}{8} = \sum_{k=1}^\infty \frac{1}{(2k-1)^2} $ – 2017-01-30

1 Answers 1

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Notice that,

$$\ln(\Gamma(x+1))=\ln(x\Gamma(x))=\ln(x)+\ln(\Gamma(x))$$

$$\ln(\Gamma(x+1))=\ln(x)+\ln(\Gamma(x))\tag0$$

Differentiate both sides:

$$\psi^{(0)}(x+1)=\frac1{x}+\psi^{(0)}(x)$$

and differentiate again:

$$\psi^{(1)}(x+1)=\frac{-1}{x^2}+\psi^{(1)}(x)\tag1$$

Thus, we may prove your statement by induction. It is true for $m=1$, and if it is true for $n$, then it is true for $n+1:$

$$\begin{align}\sum_{k=1}^{n+1}\frac1{(2k-1)^2}&=\frac{\pi^2}8-\frac14\psi^{(1)}\left(n+\frac12\right)+\frac1{(2n+1)^2}\\&=\frac{\pi^2}8-\frac14\left(\psi\left(n+1+\frac12\right)+\frac1{\left(n+\frac12\right)^2}\right)+\frac1{(2n+1)^2}\tag1\\&=\frac{\pi^2}8-\frac14\psi^{(1)}\left(n+1+\frac12\right)-\frac1{(2n+1)^2}+\frac1{(2n+1)^2}\\&=\frac{\pi^2}8-\frac14\psi^{(1)}\left(n+1+\frac12\right)\end{align}$$

QED

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    Good answer. I think those tags are unnecessary though. – 2017-01-30
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    @S.C.B. Pft, I used plenty of calculus in my answer e.g. derivatives. – 2017-01-30
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    But is the $\text{(1)}$, $\text{(0)}$ really that helpful? I personally don't think so, though it's your answer. – 2017-01-30
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    The one written at the farmost right I mean. – 2017-01-30
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    @S.C.B. Well, I used $(0)$ to get $(1)$ to prove induction, which is the entirety of what my answer aimed for. – 2017-01-30
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    @user1952009 Oh, well that came with "this is true for $m=1$", or at least an equivalent statement. – 2017-01-30
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    @user1952009 You sure you wouldn't rather just link to wikipedia or something like that? And why $\psi^{(1)}(1/2)$ instead of $\psi^{(1)}(3/2)$ – 2017-01-30
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    So $\psi^{(1)}(x+1)=\frac{-1}{x^2}+\psi^{(1)}(x)$ is really a proof that $\sum_{k=1}^{m} \frac{1}{(2k-1)^2}=4 \psi^{(1)}(1/2)-\frac{1}{4}\psi^{(1)}\left(m+\frac{1}{2}\right)$ but no clue for $4 \psi^{(1)}(1/2) = \frac{8}{\pi^2}$. From $\lim_{m \to \infty} \psi^{(1)}(m+1/2) = 0$ we obtain $4 \psi^{(1)}(1/2) = \sum_{k=1}^{\infty} \frac{1}{(2k-1)^2}$, but not $ \sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} = \frac{8}{\pi^2}$, which needs something like $\frac{\pi^2}{\sin^2(\pi z)} = \sum_{n=-\infty}^\infty \frac{1}{(z+n)^2}$ – 2017-01-30
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    @user1952009 -_- Or... we could start with$$\Gamma(1-x)\Gamma(x)=\frac\pi{\sin(\pi x)}$$Take the log and differentiate twice. – 2017-01-30
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    @SimplyBeautifulArt There are some good chances you will use $\frac{\pi^2}{\sin^2(\pi z)} = \sum_{n=-\infty}^\infty \frac{1}{(z+n)^2}$ for proving it – 2017-01-30
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    @user1952009 But that is, likewise, easily derived via the comment above. Probably my method mentioned above your comment requires less knowledge. – 2017-01-30
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    What do you mean "easily derived" ? From what ? – 2017-01-30
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    @user1952009 I mean that it's probably easier to see that$$\ln(\Gamma(1-x))+\ln(\Gamma(x))=\ln(\pi)-\ln(\sin(\pi x))$$and to differentiate both sides twice, easily revealing the value of $\psi^{(1)}(1/2)$ or $\psi^{(n)}(1/2)$ for arbitrary whole $n$. – 2017-01-30
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    The way I know to prove it is from $\frac{\pi^2}{\sin^2(\pi z)} = \sum_{n=-\infty}^\infty \frac{1}{(z+n)^2}$ or $cotan(z) = \sum_n \frac{1}{z+n}$, [using the Liouville theorem](http://math.stackexchange.com/a/1931892/276986) – 2017-01-30
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    @user1952009 Congrats for you then. – 2017-01-30
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    Come on. That's what I meant : the OP might be lost if you don't tell him what you can prove and what you are assuming – 2017-01-30
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    @user1952009 I think the OP is more likely to get lost reading your comments IMO, and I'm sure the OP is more than capable of *asking a few questions* himself if something is not clear. i.e. don't make assumptions about people. – 2017-01-30
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    @SimplyBeautifulArt Stop being insulting... As I said the point is that proving those things is not trivial. The obvious assumption is that it is not trivial for you neither, nor the OP. – 2017-01-30
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    @user1952009 O_O I'm pretty sure it was fairly trivial for me... >_> – 2017-01-30
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    Sure, the Basel problem is trivial. I downvoted your answer. – 2017-01-30
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    @user1952009 As you wish then good sir. – 2017-01-30
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    thank's for your solution!Very good!! – 2017-01-30
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    :-) no problem @IsraelMeirelesChrisostomo – 2017-01-30