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I came across the following sum: $$\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}\frac{4^k}{{2k \choose k}}$$


I thought that this can be evaluated using the expansion of $\dfrac{\sin^{-1}x}{\sqrt{1-x^2}}$ but I couldn't make any use of it. Then I tried to use the following: $$\frac{1}{(2k+1)}\frac{1}{{2k \choose k}}=\frac{\Gamma(k+1)\Gamma(k+1)}{\Gamma(2k+2)}=\int_0^1 x^k(1-x)^k\,dx$$ but this didn't help either and now I am stuck.

Any help is greatly appreciated. Thanks!


EDIT:

The sum originated from the following definite integral:

$$\int_0^{\pi/2}\tan^{-1}(\sin x)\,dx$$

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$$ \begin{align} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}\frac{4^k}{{2k \choose k}} &= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{2k+1} \int_{0}^{\pi /2} \sin^{2k+1} (x) \, dx \\ &= \int_{0}^{\pi /2} \sum_{n=0}^{\infty} \frac{(-1)^{k} \sin^{2k+1} (x)}{2k+1} \, dx\\ &= \int_{0}^{\pi /2} \arctan (\sin x) \, dx \\ &= \int_{0}^{1} \frac{\arctan t}{\sqrt{1-t^{2}}} \, dt \end{align}$$

Let $ \displaystyle I(a) = \int_{0}^{1} \frac{\arctan (at)}{\sqrt{1-t^{2}}} \ dt$.

Then differentiating under the integral sign,

$$ \begin{align} I'(a) &= \int_{0}^{1} \frac{t}{(1+a^{2}t^{2})\sqrt{1-t^{2}}} \, dt \\ &= \int_{0}^{1} \frac{1}{[1+a^{2}(1-u^{2})]u} \, u \, du \\ &= \frac{1}{1+a^{2}} \int_{0}^{1} \frac{1}{1-\left( \frac{au}{\sqrt{1+a^{2}}}\right)^{2}} \, du \\ &= \frac{1}{a \sqrt{1+a^{2}}} \text{arctanh} \left( \frac{a}{\sqrt{1+a^{2}}} \right) \\ &= \frac{1}{a\sqrt{1+a^{2}}} \frac{1}{2} \ln \Big((a+\sqrt{1+a^{2}})^{2} \Big) \\ &= \frac{1}{a \sqrt{1+a^{2}}} \ln \left( a+ \sqrt{1+a^{2}} \right) \\ &= \frac{1}{a \sqrt{1+a^{2}}} \text{arcsinh}(a) . \end{align}$$

And then integrating back,

$$ \begin{align} I(1)-I(0) = I(1) &= \int_{0}^{1} \frac{\text{arcsinh}(a)}{a \sqrt{1+a^{2}}} \, da \\ &= - \text{arcsinh}(a) \text{arcsinh}(\frac{1}{a}) \Bigg|^{1}_{0} + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \, da \\ &= - \text{arcsinh}^{2}(1) + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \, da \\ &= - \ln^{2}(1+\sqrt{2}) + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \, da . \end{align}$$

Now let $ \displaystyle w = \frac{1}{a}$.

Then

$$ I(1) = - \ln^{2}(1+\sqrt{2}) + \int_{1}^{\infty} \frac{\text{arcsinh}(w)}{w \sqrt{1+w^{2}}} \, dw$$

$$ = - \ln^{2}(1+\sqrt{2}) + I(\infty) - I(1) .$$

Therefore,

$$ \begin{align} I(1) &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{I(\infty)}{2} \\ &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi}{4} \int_{0}^{1} \frac{1}{\sqrt{1-t^{2}}} \, dt \\ &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi^{2}}{8} . \end{align}$$

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  • $\begingroup$ Awesome! Never thought about evaluating it this way, thank you Random Variable! :D $\endgroup$ – Pranav Arora Jul 4 '14 at 17:46
  • $\begingroup$ @PranavArora You're welcome. $\endgroup$ – Random Variable Jul 4 '14 at 17:58
  • $\begingroup$ @RandomVariable: (+1) I see that your answer does actually answer this related question since you show the sum equal to $\int_0^{\pi/2}\arctan(\sin(x))\,\mathrm{d}x$, which I showed in the first half of my answer to that question (but never posted since that question was closed as a duplicate of this one). I moved the second half of my answer to that question here. $\endgroup$ – robjohn Dec 5 '14 at 7:46
  • $\begingroup$ @robjohn and RV: How about this one? Does it have a closed-form? $\endgroup$ – Venus Dec 5 '14 at 16:04
  • $\begingroup$ RV: How did you get this one $$\int_{1}^{\infty} \frac{\text{arcsinh}(w)}{w \sqrt{1+w^{2}}}=I(\infty)-I(1)$$I only know $$\int_{1}^{\infty} \frac{\text{arcsinh}(w)}{w \sqrt{1+w^{2}}}=\int_{0}^{\infty} \frac{\text{arcsinh}(w)}{w \sqrt{1+w^{2}}}-\int_{0}^{1} \frac{\text{arcsinh}(w)}{w \sqrt{1+w^{2}}}$$ $\endgroup$ – Venus Dec 5 '14 at 16:19
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Determine a Related Generating Function

Using the Beta function, we get the following identity: $$ \frac1{\binom{2n}{n}}=(2n+1)\int_0^1t^n(1-t)^n\mathrm{d}t\tag{1} $$ Thus, $$ \begin{align} \sum_{n=0}^\infty\frac{(-4)^nx^{2n}}{(2n+1)\binom{2n}{n}} &=\int_0^1\frac1{1+4x^2t(1-t)}\mathrm{d}t\tag{2a}\\ &=\int_0^1\frac1{1+x^2-x^2(2t-1)^2}\mathrm{d}t\tag{2b}\\ &=\frac1{1+x^2}\int_0^1\frac1{1-\frac{x^2}{1+x^2}(2t-1)^2}\mathrm{d}t\tag{2c}\\ &=\frac1{1+x^2}\int_{-1}^1\frac1{1-\frac{x^2}{1+x^2}t^2}\frac12\mathrm{d}t\tag{2d}\\ &=\frac1{2x\sqrt{1+x^2}}\int_{-x/\sqrt{1+x^2}}^{x/\sqrt{1+x^2}}\frac1{1-t^2}\mathrm{d}t\tag{2e}\\ &=\frac1{x\sqrt{1+x^2}}\mathrm{arctanh}\left(\frac{x}{\sqrt{1+x^2}}\right)\tag{2f}\\ &=\frac1{x\sqrt{1+x^2}}\mathrm{arcsinh}(x)\tag{2g} \end{align} $$ Explanation:
$\text{(2a)}$: multiply $(1)$ by $\frac{(-4)^nx^{2n}}{2n+1}$ and sum using the formula for the sum of a geometric series
$\text{(2b)}$: rearrange the denominator of the integrand
$\text{(2c)}$: divide numerator and denominator by $1+x^2$
$\text{(2d)}$: substitute $t\mapsto(t+1)/2$
$\text{(2e)}$: substitute $t\mapsto t\sqrt{1+x^2}\,/x$
$\text{(2f)}$: integrate
$\text{(2g)}$: $\tanh(x)=\left.\sinh(x)\middle/\sqrt{1+\sinh^2(x)}\right.$


Integrate the Generating Function to Evaluate the Sum $$ \begin{align} \sum_{n=0}^\infty\frac{(-4)^n}{(2n+1)^2\binom{2n}{n}} &=\int_0^1\frac1{x\sqrt{1+x^2}}\mathrm{arcsinh}(x)\,\mathrm{d}x\tag{3a}\\ &=-\int_0^1\mathrm{arcsinh}(x)\frac1{\sqrt{\vphantom{\big|}1+1/x^2}}\mathrm{d}\frac1x\tag{3b}\\ &=-\int_0^1\mathrm{arcsinh}(x)\,\mathrm{d}\,\mathrm{arcsinh}\left(\frac1x\right)\tag{3c}\\ &=-\,\mathrm{arcsinh}^2(1)+\int_0^1\mathrm{arcsinh}\left(\frac1x\right)\,\mathrm{d}\,\mathrm{arcsinh}(x)\tag{3d}\\ &=-\,\mathrm{arcsinh}^2(1)-\int_1^\infty\mathrm{arcsinh}(x)\,\mathrm{d}\,\mathrm{arcsinh}\left(\frac1x\right)\tag{3e}\\ &=-\,\mathrm{arcsinh}^2(1)+\int_1^\infty\frac1{x\sqrt{1+x^2}}\mathrm{arcsinh}(x)\,\mathrm{d}x\tag{3f}\\ &=-\frac12\,\mathrm{arcsinh}^2(1)+\frac12\int_0^\infty\frac1{x\sqrt{1+x^2}}\mathrm{arcsinh}(x)\,\mathrm{d}x\tag{3g}\\ &=-\frac12\,\mathrm{arcsinh}^2(1)+\frac12\int_0^\infty\frac{t\,\mathrm{d}t}{\sinh(t)}\tag{3h} \end{align} $$ Explanation:
$\text{(3a)}$: integrate $\text{(2g)}$
$\text{(3b)}$: pull a factor of $x$ from the square root
$\text{(3c)}$: prepare to integrate by parts
$\text{(3d)}$: integrate by parts
$\text{(3e)}$: substitute $x\mapsto1/x$
$\text{(3f)}$: $\mathrm{d}\,\mathrm{arcsinh}\left(\frac1x\right)=-\frac1{x\sqrt{1+x^2}}\,\mathrm{d}x$
$\text{(3g)}$: average $\text{(3a)}$ and $\text{(3f)}$
$\text{(3h)}$: substitute $x=\sinh(t)$

Expand $\frac1{\sinh(x)}$ as a series in $e^{-kt}$: $$ \begin{align} \int_0^\infty\frac{t\,\mathrm{dt}}{\sinh(t)} &=\int_0^\infty\sum_{k=0}^\infty2t\,e^{-(2k+1)t}\,\mathrm{d}t\\ &=\sum_{k=0}^\infty\frac2{(2k+1)^2}\\ &=\frac{\pi^2}4\tag{4} \end{align} $$ Combining $(3)$ and $(4)$ yields $$ \begin{align} \sum_{n=0}^\infty\frac{(-4)^n}{(2n+1)^2\binom{2n}{n}} &=\frac{\pi^2}8-\frac12\mathrm{arcsinh}^2(1)\tag{5} \end{align} $$

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Here is a little something. Along the same line as your thoughts.

$$\sum_{k=0}^{\infty}\frac{(-1)^{k}4^{k}}{(2k+1)^{2}\binom{2k}{k}}$$

$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}4^{k}k!k!}{(2k+1)(2k+1)(2k)!}$$

$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}4^{k}k!k!}{(2k+1)(2k+1)!}$$

$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}4^{k}\Gamma(k+1)\Gamma(k+1)}{(2k+1)\Gamma(2k+2)}$$

$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}4^{k}\beta(k+1,k+1)}{2k+1}$$

$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}\int_{0}^{1}(4x(1-x))^{n}dx$$

$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}\int_{0}^{1}(1-(2x-1)^{2})^{n}dx$$

Let $t=2x-1$

$$1/2\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}\int_{-1}^{1}(1-t^{2})^{n}dt$$

$$=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}\int_{0}^{1}(1-t^{2})^{n}dt$$

let $t=\sin(x)$

$$\int_{0}^{\frac{\pi}{2}}\cos^{2}(x)\sum_{k=0}^{\infty}\frac{(-1)^{k}\cos^{2k-1}}{2k+1}dx$$

Note the series is arctan at cos(x)

$$=\int_{0}^{\frac{\pi}{2}}\tan^{-1}(\cos(x))dx$$

Parts:

$u=\tan^{-1}(\cos(x)), \;\ dv=dx, \;\ du=\frac{-\sin(x)}{1+\cos^{2}(x)}dx, \;\ v=x$

$$0+\int_{0}^{\frac{\pi}{2}}\frac{x\sin(x)}{1+\cos^{2}(x)}dx$$

This is as far as I have gotten for now. It seems to me this may have been done on the site at some point if we can find it.

See here for the evaluation of this integral. SOS done it sometime back.

http://sos440.tistory.com/83

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  • $\begingroup$ That's the integral from where I got the sum. I was trying to solve that integral and got stuck at the sum. $\endgroup$ – Pranav Arora Jul 4 '14 at 9:01
  • $\begingroup$ Oh, OK. I did not realize that. Well, that sin cos integral looks familiar. I bet it is on the site somewhere. $\endgroup$ – Cody Jul 4 '14 at 9:25
  • $\begingroup$ I added a link to the eval of that last integral. SOS done it on his blog. Something very similar anyway. $\endgroup$ – Cody Jul 4 '14 at 9:45
  • $\begingroup$ Thank you for the link Cody, SOS evaluated it nicely. :) $\endgroup$ – Pranav Arora Jul 4 '14 at 11:41
  • $\begingroup$ @Aryabhata: I really did not know if that was a common practice. I am sorry, I will be more careful next time. :) $\endgroup$ – Pranav Arora Jul 4 '14 at 17:45

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