# Finding the sum of convergent series with an integral

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.

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*}
• 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: $1-x^3 + \sum_{i=2} ^{\infty} x^{3i}(1-x^3).$ – user249018 Jun 14 at 13:17
• Yes, should be $6i$. Just edited it. – Chrystomath Jun 14 at 13:21
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 $$0 by the same argument as you write, and this is true for such $$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$$.