How do I evaluate this sum: :$\sum_{n=3}^{\infty} \frac{1}{1-n+n^2-\cdots+(-n)^k}$ and $k$ is positive integer?

Question

Wolfram alpha showed after some calculations for evaluation of this series :

$\sum_{n=3}^{\infty} \frac{1}{1-n+n^2-\cdots+(-n)^k}$ for example for $k=10$ ,

I have got this result which it close to $0$.

My question here is : : How do I evaluate this sum: :$$\sum_{n=3}^{\infty} \frac{1}{1-n+n^2-\cdots+(-n)^k}$$ ?

Note: I exclude the singularity points just i would like to Know how do i evaluate it

Thank you for any help

Answer 1

Hint: The denominator can be written as $$\sum_{j=0}^k (-n)^j$$ And this has a closed form: $$\frac{1-(-n)^{k+1}}{n+1} = \frac{n(-n)^k+1}{n+1}$$
From here there is no way to clean up the sum a lot more with elementary methods. More than likely you will require special functions.

Edit: In case the OP wants a little more, if we sum the reciprocal of the above fraction from $n=3$ to $n=m$ and do a bit of rearranging we can write the sum as: $$\sum_{n=3}^m \frac{1}{\sum_{j=0}^k (-n)^j} = \frac{(-1)^{k}\bigg(\zeta(k,3)-\zeta(k,m+1)\bigg)}{k+1}$$

Answer 2

The denominator is a geometric series which we have a formula for. So the summand is the reciprocal of that formula:

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

Assuming convergence, you can split this into two sums.

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

These are going to be some sort of Digamma functions.