# Is there a closed form for $\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2?$

We have

$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot{2n-1\over 2n+1}=4-\pi\tag1$$

I would like to know if there exist a closed form for

$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2 =\,??\tag2$$

I was able to roughly estimate it as $\approx\sqrt{8+2\pi}$ but it is not the closed form.

How can we find the closd form for $(2)?$

• wolframalpha.com/input/… – Ahmed S. Attaalla Aug 21 '17 at 21:42
• I wouldnt be surprised if a relatively simple closed form exists, but I also wouldnt be too surprised if one doesnt – Brevan Ellefsen Aug 21 '17 at 21:46
• From Mathematica: $$2\,_3F_2\left(\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2};\frac{1}{2}\right)-\frac{4}{9}\,_3F_2\left(\frac{3}{2},2,2;\frac{5}{2},\frac{5}{2};\frac{1}{2}\right)+\frac{4}{9}\,_4F_3\left(\frac{3}{2},2,2,2;1,\frac{5}{2},\frac{5}{2};\frac{1}{2}\right)$$ – Brevan Ellefsen Aug 21 '17 at 21:50
• Inverse symbolic calculators find nothing, so it's unlikely there is a closed expression. – Professor Vector Aug 21 '17 at 22:06
• Do you have a reason to believe that a closed from exists or are you just hoping? – Michael Burr Aug 21 '17 at 22:14

Let us denote: $$S_p := \sum\limits_{n=0}^\infty \frac{2^{n+1}}{\binom{2 n}{n}} \cdot \frac{1}{(2n+1)^p}$$ then the sum in question is just equal $S_0 - 4 S_1+4 S_2$. Now we have: \begin{eqnarray} S_0 &=& \sum\limits_{n=0}^\infty 2^{n+1} \cdot \underbrace{(2n+1)}_{\left. d_\theta \theta^{2n+1} \right|_{\theta=1}} \cdot \int\limits_0^1 t^n (1-t)^n dt \\ &=& \left. d_\theta 2 \theta \int\limits_0^1 \frac{1}{1-2 \theta^2 t (1-t)} dt \right|_{\theta=1} \\ &=& \left. d_\theta \frac{4 \arctan(\frac{\theta}{\sqrt{2-\theta^2}})}{\sqrt{2-\theta^2}} \right|_{\theta=1} = 4+\pi \end{eqnarray} We compute $S_1$ in exactly the same way. We have: \begin{eqnarray} S_1 &=& \sum\limits_{n=0}^\infty 2^{n+1} \int\limits_0^1 t^n (1-t)^n dt \\ &=& 2 \int\limits_0^1 \frac{1}{1-2 t(1-t) } dt = \frac{4 \arctan(1)}{1} = \pi \end{eqnarray} Now comes a little harder task. We have: \begin{eqnarray} S_2 &=& \sum\limits_{n=0}^\infty 2^{n+1} \cdot \underbrace{\frac{1}{(2n+1)}}_{\int\limits_0^1 \theta^{2 n} dt}\cdot \int\limits_0^1 t^n (1-t)^n dt \\ &=& 2 \int\limits_0^1 \int\limits_0^1 \frac{1}{1-2 \theta^2 t (1-t)} dt d\theta \\ &=& \int\limits_0^1 \frac{4 \arctan(\frac{\theta}{\sqrt{2-\theta^2}})}{\theta \sqrt{2-\theta^2}} d \theta \\ &\underbrace{=}_{u = \theta/\sqrt{2-\theta^2}}& 2 \sqrt{2} \int\limits_0^1 \frac{\arctan(u)}{u \sqrt{1+u^2}} du \\ &\underbrace{=}_{v=\arctan(u)}&2 \sqrt{2} \int\limits_0^{\frac{\pi}{4}} \frac{v}{\sin(v)} dv \\ &=& 2 \sqrt{2} \left.\left[v \left( \log(1-e^{\imath \cdot v}) - \log(1+e^{\imath \cdot v}) \right) + \imath \left( Li_2(-e^{\imath \cdot v}) - Li_2(e^{\imath \cdot v})\right)\right]\right|_{v=0}^{v=\pi/4} \\ &=& \sqrt{2} \frac{\pi}{2} \left( \left(\log(1-e^{\frac{\imath \pi}{4}}) - \log(1+e^{\frac{\imath \pi}{4}})\right) + \imath \frac{4}{\pi} \left( Li_2(-e^{\frac{\imath \pi}{4}}) - Li_2(e^{\frac{\imath \pi}{4}})\right)+\imath \pi\right)\\ &=& \frac{1}{16} \left( \zeta(2,\frac{1}{8})+\zeta(2,\frac{3}{8})-\zeta(2,\frac{5}{8})-\zeta(2,\frac{7}{8})\right) + \frac{\pi}{\sqrt{2}} \log(-1+\sqrt{2}) \end{eqnarray} In the last line we used the following identities: \begin{eqnarray} &&\log\left[ 1- e^{\imath \pi/4}\right] - \log\left[ 1+ e^{\imath \pi/4}\right] = \log\left[\sqrt{2}-1\right] - \imath \frac{\pi}{2}\\ &&Li_2(-e^{\frac{\imath \pi}{4}}) - Li_2(e^{\frac{\imath \pi}{4}}) = -\frac{1}{32 \sqrt{2}} \left( \right. \\ &&\zeta(2,\frac{1}{8}) - \zeta(2,\frac{3}{8}) - \zeta(2,\frac{5}{8})+\zeta(2,\frac{7}{8}) + \imath \left( \zeta(2,\frac{1}{8}) + \zeta(2,\frac{3}{8}) - \zeta(2,\frac{5}{8})-\zeta(2,\frac{7}{8})\right)\left.\right) \end{eqnarray} Here we only note that for generic $p\ge 1$ we have: \begin{eqnarray} &&S_{p+1} = -4 \imath \sqrt{2} \sum\limits_{t=1}^p \sum\limits_{s=0}^{p-t} \\ && \frac{(-1)^s}{2^{p-1} (p-s-t)! s! (t-1)!} (\log(-2))^{p-s-t} 2^{s+t-1} \int\limits_1^{\exp(\imath \pi/4)} \frac{[\log(z^2-1)]^s [\log(z)]^t}{z^2-1} d z \end{eqnarray} In general we also have: \begin{eqnarray} &&S_{p+1}(x) = -\imath 8 x \sum\limits_{t=1}^p \sum\limits_{s=0}^{p-t} \frac{(-1)^s}{(p-s-t)!s! (t-1)!} \cdot \left( \imath \frac{\pi}{2}+\log(2 x)\right)^{p-s-t} \cdot \\ &&\int\limits_1^{\exp(\imath \arcsin(x))} \frac{[\log(z^2-1)]^s [\log(z)]^t}{z^2-1} dz \end{eqnarray} where $$S_{p+1}(x) := \sum\limits_{n=0}^\infty \frac{(2 x)^{2n+2}}{\binom{2n}{n}} \cdot \frac{1}{(2n+1)^{p+1}}$$ It is still not clear whether the result reduces to polylogarithms only since for the time being we are unable to find the integrals in question.

• nice +1, it would be nice to rewrite the endresult in a manifestly real fashion – tired Aug 22 '17 at 18:06
• +1 Using your answer, we then get, $$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2=4-3\pi-2\pi^2+2\sqrt2\,\pi\ln(-1+\sqrt2)+\tfrac12\psi^{(1)}\big(\tfrac18\big)+\tfrac12\psi^{(1)}\big(\tfrac38\big)$$ with $\psi^{(n)}(z)$ as PolyGamma[n,z] in Mathematica syntax. – Tito Piezas III Aug 23 '17 at 15:21
• @Przemo: Nice derivation (+1). Very nice representation in the last line. – Markus Scheuer Aug 24 '17 at 10:28

This is not an answer at all.

Just out of curiosity, I had (using a CAS) a look at $$S_{a,b}=\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\,\frac{(2n-1)^a}{(2n+1)^b}\qquad \qquad a\geq 1\qquad b\geq 1$$ and what it seems is that, as soon as $b>1$, the result is given as linear combinations of hypergeometric functions as Brevan Ellefsen already commented.

However, what looks to be interesting is the case $b=1$ for which very simple expressions are obtained just as in $(1)$. $$\left( \begin{array}{cc} a & S_{a,1} \\ 1 & 4-\pi \\ 2 & 5 \pi \\ 3 & 68+11 \pi \\ 4 & 512+185 \pi \\ 5 & 7300+2279 \pi \\ 6 & 116224+37085 \pi \\ 7 & 2204868+701651 \pi \\ 8 & 48073728+15302705 \pi \\ 9 & 1186130180+377556239 \pi \\ 10 & 32669570048+10399048565 \pi \end{array} \right)$$

By the way, for the approximation of you result for $a=b=2$, the simple $\frac{5000}{1323}$, $\frac{43613}{11540}$, $\frac{92226}{24403}$ correspond to very small relative errors.

• @NightOfPower. You are welcome ! The problem is interesting and I had fun with this case $b=1$. Cheers. – Claude Leibovici Aug 22 '17 at 9:51

A somewhat easier way is the following:

$$\sum_{n=0}^{\infty}\frac{(2x)^{2n}}{\binom{2n}{n}}=\frac{1}{1-x^2}+\frac{x\arcsin x}{(1-x^2)^{\frac{3}{2}}}$$

Now, all we need is to integrate this expression twice and to differentiate it twice with respect to $x$.

I skip these simple procedures and write down only the end result.

$$\sum_{n=0}^{\infty}\frac{(2)^{2n}(x)^{2n-2}}{\binom{2n}{n}}\left (\frac{2n-1}{2n+1}\right )^2=$$

$$=\frac{1}{x^2(1-x^2)}-\frac{4-5x^2}{x^3(1-x^2)^{\frac{3}{2}}}\arcsin x+\frac{2\arcsin^2 x}{x^4}+\frac{2}{x^3}\int_{0}^{x}\left (\frac{\arcsin t}{t}\right )^2dt$$

To get the original sum, wich we denote as $S$, we evaluate this expression at $x=\frac{1}{\sqrt{2}}$

So the final result:

$$S=\frac{\pi^2}{2}-3\pi+4+4\sqrt{2}\int_{0}^{\frac{1}{\sqrt{2}}}\left (\frac{\arcsin t}{t}\right )^2dt$$

• \begin{eqnarray} 4 \sqrt{2} \int\limits_0^{1/\sqrt{2}} (\arcsin(t)/t)^2 dt &=& -\pi^2/2 + 8 \sqrt{2} \int\limits_0^{\pi/4} u/\sin(u) du \\ -5/2 \pi^2 + 1/2 \left( \psi^{(1)}(1/8) + \psi^{(1)}(3/8)\right) + 2 \pi \sqrt{2} \log(-1+\sqrt{2})\end{eqnarray} which gives $$S = 4- 3 \pi - 2 \pi^2 + 2 \pi \sqrt{2} \log(-1+\sqrt{2}) + 1/2 \left( \psi^{(1)}(1/8) + \psi^{(1)}(3/8)\right)$$ as given in the answers above. – Przemo Aug 31 '17 at 11:34
• Unfortunately the second formula in your answer is wrong.I checked it numerically. Besides if you input $x=1/\sqrt{2}$ into the rhs of that formula you do not get the final result (the coefficient at $\pi$ is $-3/\sqrt{2}$ instead of $-3$). Can you please fix the typo. I wanted to use your formula to find generalizations of the sum in question. – Przemo Sep 1 '17 at 16:10
• @Przemo there was a typo in the second formula. But the final formula is still correct. – Martin Gales Sep 2 '17 at 14:42
• Yes, there was just a typo in your formula . It didn't take me long to figure it out and fix it. By the way those calculations are very addictive;-). The generalized version of our sum (see my answer below) leads inevitably to an integral with two logarithms in it. I am not quite sure if the result will be reducible to polylogarithms only. Even if it will there will be a plethora of different terms anyway. – Przemo Sep 4 '17 at 11:31

This is not an answer to this question but an attempt to answer a generalized question. Our generalization consists in replacing the power two in the last term in the parentheses by a power three.Now, by starting from the (corrected--the term $(1-x^2)$ in the denominator in front of the first power of the arc sine should be raised to the power $3/2$ and not as it is now to the power one) second formula given in the answer by Martin Gales and then by multiplying the right hand side of that formula by $x^2$ and then integrating over $x$ we have derived the following formula: \begin{eqnarray} &&\sum\limits_{n=0}^\infty \frac{2^{2 n} x^{2n+1}}{\binom{2 n}{n}} \cdot \frac{(2n-1)^2}{(2n+1)^3}=\\ &&\left(4 \log(1+\sqrt{1-x^2})-4 \log(x)+ \frac{1}{\sqrt{1-x^2}}\right) \arcsin(x) + \\ &&4 \imath \left( \log\left[\frac{\left(x-\imath(1-\sqrt{1-x^2})\right)^2}{2-2 \sqrt{1-x^2}}\right] \log\left[ \frac{1-\sqrt{1-x^2}}{x}\right]-\right.\\ &&\left.Li_2[(-\imath) \frac{1-\sqrt{1-x^2}}{x} ]+Li_2[(\imath) \frac{1-\sqrt{1-x^2}}{x})]\right)+\\ &&2 \int\limits_0^x \left(\frac{\arcsin(t)}{t}\right)^2 \left(1+\log(\frac{x}{t})\right)dt =\\ &&-4 i \text{Li}_2\left(-\frac{i \left(1-\sqrt{1-x^2}\right)}{x}\right)+4 i \text{Li}_2\left(\frac{i \left(1-\sqrt{1-x^2}\right)}{x}\right)+\\ &&-8 i \text{Li}_3\left(-e^{i \sin ^{-1}(x)}\right)+8 i \text{Li}_3\left(e^{i \sin ^{-1}(x)}\right)+\\ &&\left(\text{Li}_2\left(e^{i \sin ^{-1}(x)}\right)-\text{Li}_2\left(-e^{i \sin ^{-1}(x)}\right)\right) \left(8 \sin ^{-1}(x)-4 i \log (2 i x)\right)+\\ &&\sin ^{-1}(x) \left(\frac{1}{\sqrt{1-x^2}}+4 i \log (2 i x) \cos ^{-1}\left(\frac{1}{x}\right)\right)+\\ &&4 \sin ^{-1}\left(\frac{1}{x}\right) \sin ^{-1}(x)^2+2 i \pi \log (2 i x) \cos ^{-1}(x)-14 i \zeta (3)+\\ &&8 \imath \int\limits_1^{\exp(\imath \arcsin(x))} \frac{\log(z)}{z^2-1} \log(z^2-1) dz \end{eqnarray} We will evaluate the result further and simplify it later on.