# Problem of the Week #212 - Jun 21, 2016

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#### Euge

##### MHB Global Moderator
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No one answered this week's problem. You can read my solution below.

The series evaluates to

$$\frac{1}{3} - \frac{1}{3}\log 2 + \frac{\pi}{3}\sech\left(\frac{\pi \sqrt{3}}{2}\right)$$

Indeed, since

$$\frac{n^2}{n^3 + 1} = \frac{1}{3}\left(\frac{1}{n+1} + \frac{2n-1}{n^2 + n + 1}\right)$$

then

$$\sum_{n = 1}^\infty \frac{(-1)^{n+1}n^2}{n^3 + 1} = \frac{1}{3}\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n+1} + \frac{1}{3}\sum_{n = 1}^\infty \frac{(-1)^{n+1}(2n-1)}{n^2 - n + 1}\tag{1}$$

Now

$$\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n+1} = \sum_{n = 0}^\infty \frac{(-1)^{n+1}}{n+1} - (-1) = -\log 2 + 1$$

and so the right-hand side of $(1)$ becomes

$$\frac{1}{3} - \frac{1}{3}\log 2 + \frac{1}{3}\sum_{n = 1}^\infty \frac{(-1)^{n+1}(2n-1)}{n^2 - n + 1}\tag{2}$$

We have

$$\sum_{n = 1}^\infty \frac{(-1)^{n+1}(2n-1)}{n^2 - n + 1} = \sum_{n = 1}^\infty \frac{(-1)^{n+1}(2n-1)}{(n-1/2)^2 + (3/4)} = \sum_{n = 1}^\infty (-1)^{n+1}\left(\frac{1}{n - 1/2 - i\sqrt{3}/2} + \frac{1}{n - 1/2 + i\sqrt{3}/2}\right)$$
$$= \lim_{N \to \infty} \left(\sum_{n = 1}^N \frac{(-1)^{n+1}}{n - 1/2 - i\sqrt{3}/2} + \sum_{n = 1}^N \frac{(-1)^{n+1}}{n - 1/2 + i\sqrt{3}/2}\right)$$
$$= \lim_{N \to \infty} \left(\sum_{n = -N}^{-1} \frac{(-1)^n}{n + 1/2 + i\sqrt{3}/2} + \sum_{n = 0}^{N-1} \frac{(-1)^n}{n + 1/2 + i\sqrt{3}/2}\right)$$
$$= \lim_{N\to \infty} \sum_{n = -N}^{N-1} \frac{(-1)^n}{n + 1/2 + i\sqrt{3}/2}$$
$$=\pi \sec\left(i\frac{\sqrt{3}}{2}\right)$$
$$=\pi \sech\left(\frac{\sqrt{3}}{2}\right)\tag{3}$$

Combining (1), (2), and (3), we obtain the result.

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