Question: > 
> Prove that $\sum\_{n=1}^\infty (n+1) \frac{80^n}{81^n}$ converges.
> 
> 
> 

I assume it must be done definitely by comparison or limit comparison test. But had a hard time with comparison test because of the term $(n+1)$, and when it comes to limit comparison I always get the limit as zero or infinity. Any help is appreciated.

Answer: 
Define the sequence $a\_n=(n+1)(\frac{80}{81})^n$. Then

$$|\frac{a\_{n+1}}{a\_n}|=|\frac{n+1}{n}\frac{80}{81}|\rightarrow\frac{80}{81} <1\; \text{as} \; n \rightarrow \infty$$

Hence by D'Alembert's Ratio Test, the series converges
The radius of convergence of $\sum\_{n\geq 1}(n+1)z^n$ is one, since by the action of the forward/backward difference operator
$$ (1-z)\sum\_{n\geq 1}a\_n z^n =\sum\_{n\geq 1}(a\_n-a\_{n-1})z^n,\qquad (1-z)^2\sum\_{n\geq 1}a\_n z^n =\sum\_{n\geq 1}(a\_n-2a\_{n-1}+a\_{n-2})z^n$$
hence $(1-z)^2 \sum\_{n\geq 1}(n+1) z^n$ is a polynomial, namely $z(2-z)$. In particular
$$ \sum\_{n\geq 1}(n+1)\frac{80^n}{81^n} = \frac{\frac{80}{81}\left(2-\frac{80}{81}\right)}{\left(1-\frac{80}{81}\right)^2} = 6560$$
and the same can be stated through [stars and bars](https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)):
$$ \frac{1}{(1-z)^m} = \sum\_{n\geq 0}\binom{n+m-1}{m-1}z^n.$$