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I need to show that the series $\sum_{n\geq 0} \frac {(-1)^n} {3n+1}$ is convergent, with the sum, $$\sum_{n\geq 0} \frac {(-1)^n} {3n+1}=\int_0 ^1 \frac {dx}{1+x^3}.$$

I treated first the right hand side. We observe that in the interval $(0,1)$ the function $ \frac {1}{1+x^3}$ can be written as power series, $$ \frac {1}{1+x^3}=1-x^3+x^6-x^9+...$$ But this is obviously equivalent to the geometric series with the ratio $q=-x^3,$ thus $ \frac {1}{1+x^3}= \sum_{k=0} ^{\infty} q^k.$ We get, $\int_0 ^1 \frac {dx}{1+x^3}=\int _0 ^1 (\sum_{k=0} ^{\infty} q^k)\, dx.$ Since the terms under the sum are not all positive, we can not interchange the sum and integral sign unless $$\int _0 ^1 (\sum_{k=0} ^{\infty} |q^k|)\, dx < \infty .$$ One then realizes that $\sum_{k=0} ^{\infty} |q^k|\, dx=1+x^3+x^6+...=\frac {1}{1-x^3}=\frac {1}{1-q}.$ One gets (with $dq=-3x^2 dx $),$$\int _0 ^{-1} \frac {1}{1-q}\, \frac {-1}{3x^2}\,dq.$$ Unfortunately I can not solve this integral.

Can somebody help me out how to proceed by following this way?

Many thanks.

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The series is positive if you consider pairs of terms. Thus $$ \frac{1}{1+x^3}=1-x^3+x^6-x^9+\cdots=(1-x^3)+x^6(1-x^3)+\cdots$$

Hence \begin{align*}\int_0^1\frac{1}{1+x^3}dx &= \int_0^1\sum_{i=0}^\infty x^{6i}(1-x^3)dx\\ &=\sum_{i=0}^\infty\int_0^1 x^{6i}-x^{6i+3}dx\\ &=\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}\end{align*}

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  • $\begingroup$ Thanks. Is there a typo: for $i=1$ one gets $x^{3}$ even though there is no such factor ? Mybe the sum should be written as: $1-x^3 + \sum_{i=2} ^{\infty} x^{3i}(1-x^3).$ $\endgroup$ – user249018 Jun 14 at 13:17
  • $\begingroup$ Yes, should be $6i$. Just edited it. $\endgroup$ – Chrystomath Jun 14 at 13:21
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First, you need to be careful when you transform the series into integral, because of $x=1$. To be more precise, we have $$ \sum_{n\geq 0} \frac{(-1)^{n}t^{n}}{3n+1} = \int_{0}^{t} \frac{1}{1+x^{3}}dx $$ for $0<t<1$ by the same argument as you write, and this is true for such $t$ since now the geometric series converges. Now, take the limit $t\to 1$, and the Abel's limit theorem implies that the LHS converges to the desired sum.

There's another way to deal with this problem, by analyzing the partial sum. We have $$ 1 - x^{3} + x^{6} - x^{9} + \cdots + (-1)^{N} x^{3N} = \frac{1-(-x^{3})^{N+1}}{1+x^{3}} $$ and by integrating this from 0 to 1, we get $$ \sum_{n=0}^{N} \frac{(-1)^{n}}{3n+1} = \int_{0}^{1} \frac{1}{1+x^{3}}dx - \int_{0}^{1} \frac{(-x^{3})^{N+1}}{1+x^{3}}dx $$ and the last term can be bounded by $$ \int_{0}^{1}x^{3(N+1)}dx = \frac{1}{3N+4}, $$ which converges to 0 as $N\to \infty$.

To compute the integral, I think this should appear in MSE already. This note seems helpful.

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