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I would like to find the exact value of the following series:

$ \sum_{n=0}^{\infty} (-1)^{n+1}\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t$

We can easily show that the series converges using the alternating series test:

$ 0 \leq\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t \leq\frac{1}{3n+4}$

So $ \int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t \rightarrow_{n\rightarrow\infty} 0$

And $ \int_{0}^{1}\frac{t^{3n+6}}{1+t^3}\mathrm{d} t-\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t=\int_{0}^{1}\frac{t^{3n+3}(t^3-1)}{1+t^3}\mathrm{d} t\leq0 $

So the series converges.

Do you have any idea to compute the sum?

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    @Gingerjin: $ 0 \leq \int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t \leq \int_{0}^{1} t^{3n+3}\mathrm{d} t =\frac{1}{3n+4}$2012-02-23

2 Answers 2

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$\sum_{n=0}^{m-1} (-1)^{n}\int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t = \int^1_0 \frac{t^3}{1+t^3} \sum_{n=0}^{m-1} (-t^3)^{n}\mathrm{d} t$

$ = \int^1_0 \frac{t^3}{1+t^3} \frac{1 - (-t^3)^m }{1+t^3} \mathrm{d} t.$

Thus by the Dominated Convergence Theorem, $ \sum_{n=0}^{\infty} (-1)^n \int_{0}^{1}\frac{t^{3n+3}}{1+t^3}\mathrm{d} t = \lim_{m\to\infty} \int^1_0 \frac{t^3}{1+t^3} \frac{1 - (-t^3)^m }{1+t^3} \mathrm{d} t$

$ = \int^1_0 \frac{t^3}{(1+t^3)^2} \mathrm{d} t = \frac{2\sqrt{3}\pi +\log 64 -9}{54}.$

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    Great! Thank you very much!2012-02-23
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We have $ \begin{align*} \sum_{n=0}^{\infty} (-1)^{n+1} \int_{0}^{1} \frac{t^{3(n+1)}}{t^3 + 1} \; dt & = - \int_{0}^{1} \frac{t^3}{(t^3 + 1)^2} \; dt \\ &= \left[ \frac{t}{3} \frac{1}{t^3 + 1}\right]_{0}^{1} - \frac{1}{3} \int_{0}^{1} \frac{dt}{t^3 + 1} \\ &= \frac{1}{6} - \frac{1}{3} \int_{0}^{1} \left( \frac{1}{3(t+1)} -\frac{2 t-1}{6(t^2-t+1)}+\frac{1}{2(t^2-t+1)} \right) \; dt \\ &= \frac{1}{6} - \frac{1}{3} \left( \frac{1}{3} \log 2 + 0 + \frac{\pi}{3\sqrt{3}} \right) \\ &= \frac{1}{6} - \frac{1}{9} \left( \log 2 + \frac{\pi}{\sqrt{3}} \right). \end{align*}$