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I've been asked to calculate the exact form of the total sum of two alternating series, if possible. The sum of the series are the following:

$$\sum_{i=0}^\infty (-1)^i \frac{i·\log(i)}{1+i^3} $$ $$\sum_{i=0}^\infty (-1)^i \frac{i^2·\log(i)}{1+i^3} $$

I know the first one is absolute convergent, and the second one is conditionally convergent. My question is: Is there any way of calculating the exact result of any of them by hand? If so, how could I do it? I'd usually do it by partial sums, but it doesn't seem possible this time. Thanks in advance.

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HINT: Calling $$ f(x)=\sum_{i=0}^{\infty}(-1)^i \frac{i^x}{1+i^3}\hspace{0.3cm}\text{with}\hspace{0.3cm}x<3$$

the results you are looking for are $f'(1)$ and $f'(2)$. Now, $f(x)$ seems to be related with polygamma function $\psi^\nu(z)$

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