TITLE: Finding a sum of $1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$
QUESTION [8 upvotes]: I need someone to find a mistake in my soliution or maybe to solf it much more easily... I have got a sum $$1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$$ and need to evaluate it. So here's my soliution:
$$S(x)=1+\frac{x^{4}}{4\cdot2^{4}}+\frac{x^{7}}{7\cdot2^{7}}+\frac{x^{10}}{10\cdot2^{10}}+\cdots=1+\sum_{n=1}^\infty \frac{x^{3n+1}}{(3n+1)\cdot2^{3n+1}}=1+S_1(x)$$
$$(S_1(x))_x'=\left(\sum_{n=1}^\infty \frac{x^{3n+1}}{(3n+1)\cdot2^{3n+1}}\right)_x'=\sum_{n=1}^\infty \frac{x^{3n}}{2^{3n+1}}=\frac{1}{x}\sum_{n=1}^\infty \left(\frac{x}{2}\right)^{3n+1}$$
Now let's take $\frac{x}{2}=y$, then
$$S_2(y)=\sum_{n=1}^\infty y^{3n+1}=y^4+y^7+y^{10}+\cdots=\frac{y^4}{1+y^3},|y|\le1$$
$$\left(S_1(y)\right)'=\frac{1}{2y}\cdot\frac{y^4}{1-y^3}=\frac{1}{2}\cdot\frac{y^3}{1-y^3}$$
$$S_1(y)=\frac{1}{2}\int\frac{y^3}{1-y^3}dy=\frac{1}{2}\int\left(-1+\frac{1}{1-y^3}\right)dy=-\frac{1}{2}y+\frac{\sqrt{3}}{6}\arctan\left(\frac{2\left(y+\frac{1}{2}\right)}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(y+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\lvert y-1\rvert+C$$
$$S_1(x)=-\frac{1}{4}x+\frac{\sqrt{3}}{6}\arctan\left(\frac{x+1}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(\frac{x}{2}+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\left|\frac{x}{2}-1\right|+C$$
$$S_1(0)=0, C=-\frac{\sqrt{3}\pi}{36}$$
$$S(x)=-\frac{1}{4}x+\frac{\sqrt{3}}{6}\arctan\left(\frac{x+1}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(\frac{x}{2}+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\left|\frac{x}{2}-1\right|-\frac{\sqrt{3}\pi}{36}+1$$
$$S(1)=\frac{3}{4}+\frac{\sqrt{3}}{3}\arctan\left(\frac{2\sqrt{3}}{3}\right)+\frac{1}{6}\ln(7)-\frac{\sqrt{3}\pi}{36}$$
By writing this for about 2 hours I deserve extra 50 points or at least good answers... Ha ha, thanks!

REPLY [3 votes]: It's worth cleaning things up a bit:
$$1+{1\over4\cdot2^4}+{1\over7\cdot2^7}+{1\over10\cdot2^{10}}+\cdots={1\over2}+{1\over2}\left(1+{1\over4\cdot2^3}+{1\over7\cdot2^6}+{1\over10\cdot2^9}+\cdots \right)\\={1\over2}+{1\over2}f(1)$$
where
$$f(x)=x+{x^4\over4\cdot2^3}+{x^7\over7\cdot2^6}+{x^{10}\over10\cdot2^9}+\cdots$$
Now 
$$f'(x)=1+{x^3\over2^3}+{x^6\over2^6}+{x^9\over2^9}+\cdots={1\over1-\left(x\over2\right)^3}={8\over8-x^3}$$
Since $f(0)=0$, we have
$$1+{1\over4\cdot2^4}+{1\over7\cdot2^7}+{1\over10\cdot2^{10}}+\cdots={1\over2}+{1\over2}\int_0^1f'(x)dx={1\over2}+\int_0^1{4\over8-x^3}dx$$
Patient partial fractions gives
$$\int_0^1{4\over8-x^3}dx={1\over\sqrt3}\left(\arctan\left(2\over\sqrt3\right)-{\pi\over6}\right)+{1\over6}\ln7$$
and thus
$$1+{1\over4\cdot2^4}+{1\over7\cdot2^7}+{1\over10\cdot2^{10}}+\cdots={1\over2}+{\sqrt3\over3}\arctan\left(2\sqrt3\over3\right)+{1\over6}\ln7-{\pi\sqrt3\over18}\\
\approx1.016849$$
Remark:  The OP's result has $3/4$ instead of $1/2$ and $\pi\sqrt3/36$ instead of $\pi\sqrt3/18$.  Numerically it gives $1.417999...$, which is clearly too large.