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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.

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marked as duplicate by colormegone, user61527, ml0105, Eric Stucky, Martin Sleziak Apr 22 '14 at 5:56

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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.

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

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