0

Knowing that $\sum_{n=1}^\infty 3^{-n}$ converges, how does one show that $\sum_{n=1}^\infty 3^{-n} \cdot n$ converges?

I know the first series converges by the comparison test (comparing with $\sum_{n=1}^\infty 2^{-n}$), but for the second series I'm stumped.

Alhashemi
  • 3
  • 1
    Can you show that $0 \leq \frac{n}{3^n} \leq \frac{1}{2^n}$ for $n$ big enough? (or, for instance, that $n\left(\frac{2}{3}\right)^n \to 0$) – Clement C. Dec 04 '15 at 20:17
  • 1
    If you're familiar with power series, it's not very difficult to view your series as a derivative, and prove the convergence using that. – Scounged Dec 04 '15 at 20:34

1 Answers1

1

use the ratio test $$L=\lim_{n\rightarrow \infty }\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow \infty }\frac{(n+1)3^n}{n3^{n+1}}=\lim_{n\rightarrow \infty }\frac{n+1}{3n}=\lim_{n\rightarrow \infty }(\frac{1}{3}+\frac{1}{3n})=\frac{1}{3}<1$$

E.H.E
  • 23,280