What function does this Taylor Series represent? What is the function $f$ who's Taylor series is $1 - \frac{x}{4} + \frac{x^2}{7} - \frac{x^3}{10} + \cdots$ ?
I need to find the value of the series $ \sum^{\infty}_{n = 0}a_n = 1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{10} +  \cdots$ by finding $\lim_{x\rightarrow1^-} \sum^{\infty}_{n = 0}a_n x^n$. (Abel Summability)
I already did this for another problem.  I was given the series $\sum^{\infty}_{n = 0}b_n = \frac{1}{2\cdot 1} - \frac{1}{3\cdot 2} + \frac{1}{4\cdot 3} - \frac{1}{5\cdot 4} + \cdots$
I showed that $\sum^{\infty}_{n = 0}b_n x^n = \frac{\ln(1+x)}{x^2}+\frac{\ln(1+x)}{x}-\frac{1}{x}\rightarrow 2\ln(2)-1$ as $x\rightarrow1^- \Rightarrow \sum^{\infty}_{n = 0}b_n = 2\ln(2)-1$.  (This checks out numerically, as well)
However, I'm having trouble finding the functional representation of $ \sum^{\infty}_{n = 0}a_n x^n$
 A: If
$f(x)
=\sum_{k=0}^{\infty} a_k x^k
$,
then
$f_{n, j}(x)
=\sum_{k=0}^{\infty} a_{j+kn} x^{j+kn}
=\frac1{n}\sum_{i=0}^{n-1} w^{ij} f(w^i x)
$
where
$w = e^{2\pi i /n}
$.
This is known as
multisection of series.
Here is one of many
available discussions:
http://mathworld.wolfram.com/SeriesMultisection.html
Since
$f(x)
=\frac{-\ln(1-x)}{x}
=\sum_{k=0}^{\infty} \frac{x^k}{k+1}
$,
$\sum_{k=0}^{\infty} \frac{x^{3k}}{3k+1}
=f_{0, 3}(x)
=\frac1{3}\sum_{i=0}^{2}  f(w^i x)
$
where
$w = e^{2\pi i /3}
$.
Since you want
$g(x)
=\sum_{k=0}^{\infty} \frac{(-1)^kx^k}{3k+1}
$,
$g(x)
=f_{0, 3}((-x)^{1/3})
$.
A: Write the series as
$$\sum_{k=0}^{\infty} \frac{(-1)^k x^k}{3 k+1} = \sum_{k=0}^{\infty} \frac{(-1)^k \left (x^{1/3} \right )^{3 k}}{3 k+1} = x^{-1/3} \sum_{k=0}^{\infty} \frac{(-1)^k \left (x^{1/3} \right )^{3 k+1}}{3 k+1}$$
Consider
$$f(y) = \sum_{k=0}^{\infty} \frac{(-1)^k y^{3 k+1}}{3 k+1}$$
$$\implies f'(y) = \sum_{k=0}^{\infty} (-1)^k y^{3 k} = \frac1{1+y^3} $$
$$\implies f(y) = \int \frac{dy}{1+y^3} = -\frac{1}{6} \log \left(y^2-y+1\right)+\frac{1}{3} \log (y+1)+\frac{\arctan{\left(\frac{2 y-1}{\sqrt{3}}\right)}}{\sqrt{3}} +C$$
$$f(0)=0 \implies C=\frac{\pi}{6 \sqrt{3}} $$
The above integral is evaluated using a partial fraction decomposition.  Thus, the original sum is equal to

$$\sum_{k=0}^{\infty} \frac{(-1)^k x^k}{3 k+1} = \frac{\pi x^{-1/3}}{6 \sqrt{3}} + \frac{x^{-1/3}}{6} \log{\left [\frac{\left (1+x^{1/3}\right )^2}{1-x^{1/3}+x^{2/3}} \right ] } + \frac{x^{-1/3}}{\sqrt{3}} \arctan{\left(\frac{2 x^{1/3}-1}{\sqrt{3}}\right)}$$

The limit as $x \to 1^-$ of this sum is then equal to 
$$\frac{\pi}{3 \sqrt{3}} + \frac13 \log{2} $$
A: We can get a closed form with elementary functions.
Let f(x) be the function.  Define u=x^(1/3) and then g(u) = u*f(u^3).  Then g'(u) is a geometric series summing to 1/(1+u^3).  With g(0)=0 we can now set up g(u) as a definite integral.  This is carried out via partial fractions giving logarithmic and inverse tangent terms (plus a constant, as the inverse tangent term is not zero at the lower limit of the integral).  Divide that result by u and then put in u=x^(1/3) to get f(x).  The result is rather complicated as a function of x, but simplifies considerably for x=1.
If I have this right, the numerical result should be 0.8356... .
