$\sum_{n=1}^\infty (n+1) \frac{80^n}{81^n}$ converges.

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 1

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

Answer 2

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.

Answer 3

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 $$\sum_{n=1}^{\infty}(n+1)\Big(\frac{80}{81}\Big)^n=\sum_{n=0}^{\infty}(n+1)\Big(\frac{80}{81}\Big)^n-1=\frac{1}{(1-80/81)^2}-1<+\infty$$

Answer 4

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.

Answer 5

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: $$ \frac{1}{(1-z)^m} = \sum_{n\geq 0}\binom{n+m-1}{m-1}z^n.$$