# How do I evaluate this sum: :∑∞n=311−n+n2−⋯+(−n)k\sum_{n=3}^{\infty} \frac{1}{1-n+n^2-\cdots+(-n)^k} and kk is positive integer?

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

∑∞n=311−n+n2−⋯+(−n)k\sum_{n=3}^{\infty} \frac{1}{1-n+n^2-\cdots+(-n)^k} for example for k=10k=10 ,

I have got this result which it close to 00.

My question here is : : How do I evaluate this sum: :∞∑n=311−n+n2−⋯+(−n)k\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

=================

It seems that you forgot the nn term in the input to WA.
– Claude Leibovici
2 days ago

I think you want nkn^k to be (−n)k(-n)^k.
– Aweygan
2 days ago

I want the altern form in denominator
– user51189
2 days ago

thanks i would like a fixed k with altern form in denominator !!!!
– user51189
2 days ago

In my humble opinion, I think that this kind of series is very artificial. As you are young, don’t waste too much time on deadends such as this one.
– JeanMarie
2 days ago

=================

2

=================

Hint:
The denominator can be written as
k∑j=0(−n)j\sum_{j=0}^k (-n)^j
And this has a closed form:
1−(−n)k+1n+1=n(−n)k+1n+1\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=3n=3 to n=mn=m and do a bit of rearranging we can write the sum as:
m∑n=31∑kj=0(−n)j=(−1)k(ζ(k,3)−ζ(k,m+1))k+1\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}

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

∞∑n=3n+11−(−n)k+1.\sum_{n=3}^{\infty} \frac{n+1}{1-(-n)^{k+1}}.

Assuming convergence, you can split this into two sums.

∞∑n=3n1−(−n)k+1+∞∑n=311−(−n)k+1.\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.

My scratch paper has the minus sign. Stand by.
– B. Goddard
2 days ago