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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$

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A closed form by hypergeometric function is given by wolframalpha. I think you can give up on this approach. – Henry W. Feb 24 at 2:26
up vote 12 down vote accepted

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

$$\frac{\pi}{3 \sqrt{3}} + \frac13 \log{2} $$

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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}) $.

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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... .

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