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If I have \begin{eqnarray*} S = \sum_{n\geq 0}{\frac{(-1)^{n}}{3n+1}}. \end{eqnarray*}

Prove that \begin{eqnarray*} S = \int_{0}^{1}{\frac{1}{1+x^{3}}}\,\mathrm{d}x. \end{eqnarray*}

I know that serie is convergent but I don't know how to prove this, I think in some theorem of uniformly convergence and integration where \begin{eqnarray*} f(x)=\sum_{n=1}^{\infty}{f_{n}(x)} \end{eqnarray*} but I don't know how to do this. Can you give me some advice to prove it? Thank you.

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  • $\begingroup$ Look up Fubini's theorem. $\endgroup$ – K.defaoite Mar 3 at 1:05
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    $\begingroup$ Take $n\in\Bbb Z_{\ge 2}$. Then the geometric series $$\frac{1}{1+x^n}=\sum_{k\ge0}(-1)^kx^{nk}$$ converges uniformly, so we can interchange the $\sum$ and $\int$ to get $$\int_0^1\frac{dx}{1+x^n}=\int_0^1\sum_{k\ge0}(-1)^kx^{nk}dx=\sum_{k\ge0}(-1)^k\int_0^1x^{nk}dx=\sum_{k\ge0}\frac{(-1)^k}{nk+1}$$ $\endgroup$ – clathratus Mar 3 at 2:33
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We find $$\int_0^1 \frac{1}{1+x^3} dx = \int_0^1 \sum_{n \ge 0} (-x)^{3n} \ dx = \sum_{n \ge 0} (-1)^n \int_0^1 x^{3n} dx = \sum_{n \ge 0} \frac{(-1)^n}{3n+1}.$$

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  • $\begingroup$ Do you use the Taylor series of $\frac{1}{1+x^{3}}$? $\endgroup$ – angie duque Mar 3 at 1:04
  • $\begingroup$ @angieduque Geometric series. $\endgroup$ – K.defaoite Mar 3 at 1:05
  • $\begingroup$ @angie: yes, it’s the Taylor series. In this case, the Taylor series happens to be the geometric series. This answer is correct (it could have more details, but this is the correct method to prove the desired identity). $\endgroup$ – Clayton Mar 3 at 1:18
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    $\begingroup$ This is missing some crucial information. It's not always true that $\int \sum=\sum\int.$ $\endgroup$ – zhw. Mar 3 at 1:23
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The only difficulty here is justifying switching the order of integration and the summation in

$$\int_0^1 \frac{dx}{1+x^3} = \int_0^1 \sum_{n=0}^\infty (-x^3)^n\, dx = \sum_{n=0}^\infty \int_0^1 (-x^3)^n\, dx = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1}$$

We don't have uniform convergence of the series on the interval $[0,1]$ since the series diverges at $x=1$.

One way to justify is to note that

$$\sum_{n=0}^N(-x^3)^n = \frac{1 - (-x^3)^{N+1}}{1+x^3},$$

and

$$\int_0^1\sum_{n=0}^N(-x^3)^n\, dx - \int_0^1 \frac{dx}{1+x^3} = (-1)^{N+2}\int_0^1 \frac{x^{3N+3}}{1+x^3}\, dx$$

Since we can switch the integral and a finite sum , we have

$$\int_0^1\sum_{n=0}^N(-x^3)^n\, dx =\sum_{n=0}^N \int_0^1(-x^3)^n\, dx = \sum_{n=0}^N \frac{(-1)^n}{3n+1},$$

and it follows that

$$\left|\sum_{n=0}^N \frac{(-1)^n}{3n+1}- \int_0^1 \frac{dx}{1+x^3}\right| = \int_0^1 \frac{x^{3N+3}}{1+x^3}\, dx \leqslant \int_0^1 x^{3N+3}\, dx \\= \frac{1}{3N+4} \underset{N \to \infty}\longrightarrow 0$$

Thus,

$$\lim_{N \to \infty}\sum_{n=0}^N \frac{(-1)^n}{3n+1} = \int_0^1 \frac{dx}{1+x^3}$$

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