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This came from a birthday problem. I do not know much about series and I don't know which tools to use. This is similar to the problem found here, but since the coefficient differs from the exponent by two I don't know how to use a similar method. $$ \sum_{i=2}^\infty i (364/365)^{i-2} $$

Fernando Chu
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  • $\sum ia^{i-1} = \frac{d}{da}\sum a^i$ will likely help you here. Be careful with the indices. – Chinny84 May 03 '21 at 16:09
  • "...since the coeffient differs from the exponent..." Note that $\sum\limits_{i=2}^\infty i x^{i-2} = x^{-2}\cdot\sum\limits_{i=2}^\infty ix^i$ – JMoravitz May 03 '21 at 16:15

1 Answers1

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For any $x < 1$,

$\frac{d}{dx}(\sum_{i= 0}^{\infty}x^i) = \sum_{i= 1}^{\infty}ix^{i-1}$

and

$\frac{d}{dx}(\sum_{i= 0}^{\infty}x^i) = \frac{d}{dx}(\frac{1}{1-x}) = \frac{1}{(1-x)^2}$.

Hence $\sum_{i= 1}^{\infty}ix^{i-1} = \frac{1}{(1-x)^2}$

Therefore, $\sum_{i=1}^{\infty}i(\frac{364}{365})^{i-1} = 365^2 = 133225$.

Hence, $\sum_{i=1}^{\infty}i(\frac{364}{365})^{i-2} = 365^2 . \frac{365}{364} $

Therefore $\sum_{i=2}^{\infty}i(\frac{364}{365})^{i-2} = 365^2 . \frac{365}{364} - \frac{365}{364} = 133590 $