Finding similar series
Solution 1:
Well, @xxxx036 I have seen some of your questions on zeta functions they were so interesting and tough that I was finding difficulties in understanding but this disappointed me.
$$\sum_{n \ge1}\frac{(n+1)^p}{n^q} = \sum_{n\ge1}\frac {\sum_{k =0}^pn^k}{n^q} = \sum_{n\ge1}\sum_{k =1}^p\binom pkn^{k - q} = \sum_{k =1}^p\binom pk\sum_{n\ge1}n^{k - q} = \sum_{k =1}^p\binom pk\zeta(q-k)$$ Here, $q-p\ge2$
$\zeta(n) = \sum_{k\ge1}H_k^n$