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Determine the sum of

$$\sum_n^\infty \frac{k}{3^k}$$

Can someone teach me how to solve this please thanks.

Martin Sleziak
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user10024395
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  • This is called an arithmetico-geometric sequence, and it's certainly been discussed here before. – David H Apr 22 '14 at 02:30
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    At least once a week of late... Here is one of a number of related posts: http://math.stackexchange.com/questions/337937/why-sum-k-1-infty-frack2k-2 – colormegone Apr 22 '14 at 04:12
  • This is a polylogarithm. – Lucian Apr 22 '14 at 05:50
  • For the partial sums, see here: http://math.stackexchange.com/questions/11464/how-to-compute-the-formula-sum-limits-r-1d-r-cdot-2 In these questions you can find a straightforward generalization of your question: http://math.stackexchange.com/questions/441481/why-does-sum-n-0-infty-fracn2n-converge-to-2 and http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn – Martin Sleziak Apr 22 '14 at 05:55

2 Answers2

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Use summation by parts. That's integration by parts (class groans) made discrete so it's harder (another groan).

$$\sum_n^\infty \frac{k}{3^k}={-3\over 2}{k\over 3^k}(n,\infty)+{1\over 2}\sum_n^\infty \frac{1}{3^k}=\frac{3n}{2\cdot 3^n}+{1\over 2}\frac{1}{3^n(1-1/3)}=\frac{3n}{2\cdot 3^n}+{1\over 2}\frac{1}{3^{n-1}(2)}$$ and you can simplify further (more groaning).

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Generating functions can be used:

$$\frac{1}{1-x}=\sum_{k\ge 0}x^k$$

and

$$\frac{1-x^n}{1-x}=\sum_{k=0}^{n-1} x^k$$

Differentiate and multiply by x on both sides:

$$\frac{x}{{\left( 1-x \right)}^{2}}=\sum_{k\ge 0}k\, x^k\tag 1$$

and

$$x{\left(\frac{n x^{n - 1}}{x - 1} - \frac{x^{n} - 1}{{\left(x - 1\right)}^{2}}\right)} =\sum_{k=0}^{n-1}k\, x^k\tag 2$$

Subtract: $(1)-(2)$ and substitute $x=1/3$ to get the required answer.

$$\therefore \sum_{k\ge n}\frac{k}{3^k}=\frac{3}{4} \, 3^{-n} {\left(2 \, n + 1\right)}$$

gar
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