Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
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 to find the sum of this infinite series

In particular, see the answer at How to find the sum of this infinite series

Thanks all!

share|improve this question

marked as duplicate by t.b., Arturo Magidin, Aryabhata, Asaf Karagila, Sivaram May 2 '11 at 17:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4  
Very close to math.stackexchange.com/questions/30732/… . In particular, see the answer at math.stackexchange.com/questions/30732/… –  lhf May 2 '11 at 16:41
    
Thanks! If you put that in a answer form, I can give you credit for the answer easier :) –  Nitroware May 2 '11 at 16:46

3 Answers 3

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 $|x|<1$

share|improve this answer

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 $n \to \infty$.

share|improve this answer
    
I get $3 S_{n+1} - S_n = 2 S_n + \frac{n+1}{3^n}$, in the limit this is $2S = 2S$... How do you get the sum? –  quanta May 2 '11 at 16:59
1  
I get $3S_{n+1}−S_n = \sum_{k=0}^n 1/3^k$ –  Emre May 2 '11 at 17:10
    
I see! Thanks –  quanta May 2 '11 at 18:48

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 $x = 1/3$.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.