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: 
In general for $|r|<1$ we have
$$\sum\_{n=0}^{\infty}ar^{n+1}=\frac{ar}{1-r}$$
Differentiating term by term we obtain 
$$\sum\_{n=0}^{\infty} a(n+1)r^n=\frac{d}{dr}\frac{ar}{1-r}=\frac{a}{(1-r)^2}$$
In your case we have $r=80/81$ and $a=1$ and 
$$\sum\_{n=1}^{\infty}(n+1)\Big(\frac{80}{81}\Big)^n=\sum\_{n=0}^{\infty}(n+1)\Big(\frac{80}{81}\Big)^n-1=\frac{1}{(1-80/81)^2}-1<+\infty$$
Use the root test.
$$
(n+1)^{1/n}\left(\frac{80^n}{81^n}\right)^{1/n}=(n+1)^{1/n}\frac{80}{81}\to 80/81<1
$$
as $n\to\infty$. Thus the series converges.