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I need to find the sum of this series: $1, 2 \left ( 1 - \frac{1}{\sqrt{15}} \right ), 3 \left ( 1 - \frac{1}{\sqrt{15}} \right ) ^ 2, 4 \left ( 1 - \frac{1}{\sqrt{15}} \right ) ^ 3, 5 \left ( 1 - \frac{1}{\sqrt{15}} \right ) ^ 4, ... $

I easily found the formula: $\displaystyle\sum_{n=0}^{\infty} (n+1) \left ( 1 - \frac{1}{\sqrt{15}} \right ) ^ n$

but now I don't know how to find the solution without using limits. I know I could solve the limit: $\displaystyle\lim_{k\to\infty}\sum_{n=0}^{k} (n+1) \left ( 1 - \frac{1}{\sqrt{15}} \right ) ^ n$

I was wondering if there is a way to solve the sum without using limits.

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  • $\begingroup$ But you're familiar with the geometric series, yes? $\endgroup$ Oct 17, 2011 at 16:15
  • $\begingroup$ Yes, familiar enough I think... $\endgroup$
    – rubik
    Oct 17, 2011 at 16:29

1 Answer 1

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Hint: If you can do $\sum x^n$, $\frac{d}{dx}\sum x^n=\sum nx^{n-1}$ evaluated at the proper $x$ will get you there.

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    $\begingroup$ Isn't $\displaystyle\sum nx^{n-1} = \frac{1}{(1-x)^2}$? $\endgroup$
    – rubik
    Oct 17, 2011 at 16:31
  • $\begingroup$ OK, thank you, I replaced $1 - \frac{1}{15}$ instead of $1 - \frac{1}{\sqrt{15}}$, a stupid error! $\endgroup$
    – rubik
    Oct 17, 2011 at 16:34

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