Evaluating the series $\sum\limits_{n=0}^{\infty}\frac{n}{3^n}$

Question

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How to find the sum of this infinite series

Hello all, I have one last major question, where would I get started on the following question:

$$\sum_{n=0}^{\infty}\frac{n}{3^n}$$

I know it is a series (obviously), and I think it is geometric, but I have no idea as to how to start it. Does anyone have any first steps/tips as to what I could do for this?

Thanks so much in advance!

Edit: Per the first comment on my posting, by 1hf, see:

Very close to How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$

In particular, see the answer at How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$

Thanks all!

Answer 1

Hint: Observe that $$(\sum_{n=0}^\infty x^n)'=\sum_{n=0}^\infty nx^{n-1}$$ and $\sum_{n=0}^\infty x^n$ is convergent for all $|x|<1$

Answer 2

Let $S_n=\sum_{k=0}^n k/3^k$. Simplify $3 S_{n+1} - S_n$ to determine $S_n$ and then take the limit as $n \to \infty$.

Answer 3

It's (not quite) geometric. A geometric series is of the form $\sum_{n = 0}^\infty x^n$.

You are on the right track. For a geometric series, provided $|x|<1$, $$\sum_{n = 0}^\infty x^n = \frac{1}{1 - x}.$$

For convergent series, it's acceptable to differentiate term by term. This tells us, provided $|x| < 1$, $$\sum_{n = 0}^\infty nx^{n-1} = \left(\frac{1}{1-x}\right)' = \frac{1}{(1-x)^2}.$$

I claim your series is very close, but not quite, equal to this form with $x = 1/3$.