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How do we compute the following infinite series? $\sum_{k=1}^\infty k(\frac{7}{8})^{k-1}$

It arises as part of a more general problem. What is the expectations of the number of draw srequired from a uniform [0,1] distribution before you get a number that is $\epsilon$ close to 1?

Intuitively, it is $1/\epsilon$, but to show this, you would need to calculate the expectation: $E[X] = \sum_X f(X)\cdot X = \sum_{k=1}^\infty k(\frac{7}{8})^{k-1}$

Snowball
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1 Answers1

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The thing to notice is that this looks a lot like a derivative.

Let $p\in(0,1)$. Then $kp^{k-1} = \frac{d}{dp}p^{k}$

So that $\sum_{k=1}^{\infty}kp^{k-1} = \frac{d}{dp}\sum_{k=1}^{\infty}p^{k} = \frac{d}{dp}\frac{p}{1-p}$

It's now a case of some elementary differentiation and plugging in the required value of $p$

platypus22
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