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(1) Find the sum of the power series

$$\sum_{n=1}^{\infty} nx^n$$

(2) Find the sum of the series

$$\sum_{n=1}^{\infty} \frac{n}{3^n}$$

Any tips on solving the sum of series/power series?

user136088
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2 Answers2

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You can find the second sum using the first with $x = 1/3$. As for the first, for $|x| <1$ write $$ \sum_{n=1}^\infty n x^n = x \sum_{n=1}^\infty n x^{n-1} = x \sum_{n=1}^\infty \frac{d}{dx} x^n = x \frac{d}{dx} \sum_{n=1}^\infty x^n = x \frac{d}{dx} \frac{x}{1-x} = \frac{x }{(1-x)^2}. $$

user139388
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  • How do you know about the x=1/3?
    edit: nevermind!     – user136088 Apr 10 '14 at 19:08
  • Since $|1/3| < 1$ you can make the substitution $x=1/3$ in the formula for $\sum_{n=1}^\infty n x^n$ found above – user139388 Apr 10 '14 at 19:09
  • wait actually how do we get to 1/3 first? – user136088 Apr 10 '14 at 19:10
  • So, from the above deduction you have $$ \sum_{n=1}^\infty \frac{n}{3^n} = \sum_{n=1}^\infty n (1/3)^n = \frac{1/3}{(1-1/3)^2} $$ You just plug the value $x=1/3$ into the above formula! :) – user139388 Apr 10 '14 at 19:12
  • I get that part but how do we know to use x=1/3? Would it just be the first term? – user136088 Apr 10 '14 at 19:14
  • Because the sum $\sum_{n=1}^\infty \frac{n}{3^n}$ is the sum $\sum_{n=1}^\infty n x^n$ when $x=1/3$. That was the point of the first equality in my above comment. – user139388 Apr 10 '14 at 19:16
  • Hi, could you please tell me how to find the sum $\sum_{n=0}^\infty 3x^n$? It seems related, but I'm failing to do it. – Diya Nov 14 '14 at 21:23
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Hint: If you take the power series $1/(1-x) = \sum_{n=0}^\infty x^n$ and take the derivative term by term, what do you get? Also the second series is a special case of the first where $x = 1/3$.

user2566092
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