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Question: Is there a way to simplify $$\sum\limits_{k=0}^{\infty}\dfrac 1{2^{k+1}}\sum\limits_{n=0}^{\infty}\binom kn\dfrac 1{2(n+1)(3n+1)}\tag{1}$$ Into a single summation symbol? $\sum\limits_{k=0}^{\infty}\text{something}$

I inputed it into WolframAlpha and got a really complicated expression$$\sum\limits_{k=0}^{\infty}\dfrac {1-2^{k+1}+3\left(_2F_1\left[\begin{array}{c c}\frac 13,-k\\\frac 43\end{array};-1\right]\right)+3k\left(_2F_1\left[\begin{array}{c c}\frac 13,-k\\\frac 43\end{array};-1\right]\right)}{2^{k+3}(k+1)}$$ Which isn't what I really wanted because the inner sum is significantly more complex than before. Is there a way?

I'm still relatively new to this. If you have a hint, it would mean a lot if you commented it!

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Without recognizing a binomial transform, you may notice that

$$\begin{eqnarray*}\sum_{n=0}^{k}\binom{k}{n}\frac{1}{2(n+1)(3n+1)} &=& \frac{3}{4}\int_{0}^{1}\sum_{n=0}^{k}\binom{k}{n}\left(x^{3n}-x^{3n+2}\right)\,dx\\&=&\frac{3}{4}\int_{0}^{1}(1-x^2)(1+x^3)^k\,dx\end{eqnarray*}$$ hence: $$\sum_{k\geq 0}\frac{1}{2^{k+1}}\sum_{n=0}^{k}\binom{k}{n}\frac{1}{2(n+1)(3n+1)}=\frac{3}{4}\int_{0}^{1}\frac{1-x^2}{1-x^3}\,dx $$ and the last integral can be computed through partial fraction decomposition.

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  • $\begingroup$ (+1) I was going to mention this, but it appears you finished before I did. $\endgroup$ – Simply Beautiful Art Mar 11 '17 at 16:37
  • $\begingroup$ are you sure about the $\frac{1}{6}$ ? $\frac{1}{3n+1} - \frac{1}{3n+3} = \frac{2}{3(n+1)(3n+1)}$ I think it should be $\frac{3}{4}$ $\endgroup$ – Anonymous Jun 18 '17 at 0:05
  • $\begingroup$ @Anonymous: you are right, now fixing. $\endgroup$ – Jack D'Aurizio Jun 18 '17 at 0:08
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Notice that

$$\binom kn=0\forall n>k$$

Thus, it simplifies down to

$$\sum_{k=0}^\infty\frac1{2^{k+1}}\sum_{n=0}^k\binom kn\frac1{2(n+1)(3n+1)}$$

And by an inverse Euler sum, this reduces down to

$$\sum_{n=0}^\infty\frac1{2(n+1)(3n+1)}=\frac1{12}(\pi\sqrt3+9\ln(3))$$

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    $\begingroup$ Okay, I blame myself for this, but I started with $$\sum\limits_{n=0}^{\infty}\dfrac 1{2(n+1)(3n+1)}$$Performed an Euler Sum and got $(1)$. Is there a way to simplify $(1)$ into a different summation? I'm just wondering if there is a faster convergence for $$\pi=4\sqrt{3}\sum\limits_{n=0}^{\infty}\dfrac 1{2(n+1)(3n+1)}$$ $\endgroup$ – Crescendo Mar 11 '17 at 16:18
  • $\begingroup$ Lmao, thought you did. To be honest, you could just apply another Euler sum to get back to the original, but beyond that I don't know. $\endgroup$ – Simply Beautiful Art Mar 11 '17 at 16:19

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