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: 
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 finite sum can expressed by the formula $$80\ 81^{-n} \left(-80^n n-41\ 2^{4 n+1} 5^n+82\
 81^n\right)$$ and now compute the Limit for $n$ tends to infinity.