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I need to calculate $$\lim_{n \to \infty}\sum_{k=1}^{n} \frac{k}{3^k}$$ and I can't really do it. I have a feeling it's simple and that there's a simple catch but I just can't see it. If anyone could provide me with a hint to solve this, I would be grateful.

I know this question is classified as homework and that it seems like I haven't tried anything, but I have. I tried expanding this but that doesn't help at all. Geometric progression came to mind at first, but I realized it's not very useful either. Thanks.

ReliableMathBoy
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Transcendental
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2 Answers2

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Knowing that $$\sum_{k=0}^\infty n^k=\frac{1}{1-n}$$ $$\implies \frac{d}{dn}\sum_{k=0}^\infty n^k=\frac{d}{dn} \frac{1}{1-n}$$ $$\implies \sum_{k=0}^\infty kn^{k-1}=\frac{1}{(1-n)^2}$$ $$\implies \sum_{k=0}^\infty kn^k=\frac{n}{(1-n)^2}$$

Your sum is $$\sum_{k=1}^\infty k \left(\frac{1}{3}\right)^k=\sum_{k=0}^\infty k \left(\frac{1}{3}\right)^k-\sum_{k=0}^0 k \left(\frac{1}{3}\right)^k= \frac{\frac 13}{(1-\frac 13)^2}-0\,\cdot (\frac 13 )^0=\frac34$$

Teoc
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  • This solution is quite elegant but i have to say I do not recognize the methods used here. If you could please explain (or point to a Wikipedia or MathWorld article) how does, and why, this yield a correct solution and when it can be user? – Transcendental Feb 08 '15 at 09:17
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    @Transcendental In this, I have derived the formula about evaluating sums of the form $\sum_{k=1}^\infty ka^k$ from the standerd geometric sum. The steps are valid because of summation properties and a theorem regarding differentiating with sums. Because this was proven to be true, this is valid for all sums in the form $kn^k$. I will leave the task of finding the radius of convergence of this new series and its summation formula to you. – Teoc Feb 08 '15 at 18:26
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Let $S_n = \sum_{k = 1}^n k/3^k$. Then \begin{align}\frac{1}{3}S_n = \sum_{k = 1}^n \frac{k}{3^{k+1}} &= \sum_{k = 1}^n \frac{k+1}{3^{k+1}} - \sum_{k = 1}^n \frac{1}{3^{k+1}} = S_n - \frac{1}{3} + \frac{n+1}{3^{n+1}} + \frac{1}{3^2}\frac{1 - \left(\frac{1}{3}\right)^n}{1 - \frac{1}{3}}\\&= S_n - \frac{1}{3} + \frac{n+1}{3^{n+1}} +\frac{1}{6}\left[1 - \left(\frac{1}{3}\right)^n\right].\end{align} You can either isolate $S_n$, then take the limit as $n\to \infty$, or you can take the limit first to get an equation in $S_\infty := \lim_{n\to \infty} S_n$, then solve for $S_\infty$.

kobe
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