Question: > 
> Prove that $\sum\_{n=1}^\infty (n+1) \frac{80^n}{81^n}$ converges.
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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 
$$\sum\_{n=1}^{\infty}(n+1)\Big(\frac{80}{81}\Big)^n=\sum\_{n=0}^{\infty}(n+1)\Big(\frac{80}{81}\Big)^n-1=\frac{1}{(1-80/81)^2}-1<+\infty$$
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)):
$$ \frac{1}{(1-z)^m} = \sum\_{n\geq 0}\binom{n+m-1}{m-1}z^n.$$