# Calculating the total sum of alternating series

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.

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 $$\psi^\nu(z)$$