Consider this summation $$ \sum_{i=1}^{\infty}\frac{1}{i^ab^i} $$ where $a$ and $b$ are greater than $1$
It can be upper bounded by the geometric series $\sum_{i=1}^{\infty}\frac{1}{b^i}=\frac{1}{b-1}$
In addition, the summation $\sum_{i=1}^{\infty}\frac{1}{i^a}$ is called power series. It converges for $a >1$, but there is no closed form for it.
I am asking if I can evaluate the original summation exactly or not? Also, if I can get an upper bound closer than the above one or not? Since the above upper bound depends only on $b$
What you are looking at is called the Polylogarithm function: $$ \sum_{n=1}^{\infty}\frac{1}{n^ab^n}=\operatorname{Li}_a\left(\frac1b\right) $$