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Let $p$ be a prime number. How can I show that, for any positive integer $N$,

$$\sum_{i=1}^N {\frac{i}{p^i}}\le \frac{p}{(p-1)^2}?$$

I can see that

$$\sum_{i=1}^N {\frac{1}{p^i}}\lt \sum_{i=1}^\infty {\frac{1}{p^i}} = \frac{1}{p-1}$$

by the infinite sum of a geometric series, but not sure if this is useful.

Thank you.

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up vote 4 down vote accepted

Note that if $|x| <1$ $$\begin{align}\sum_{i=0}^{\infty}i\cdot x^i &=x+2x^2+3x^3+4x^4+\cdots \\ &=(x+x^2+x^3+x^4+\cdots)+(x^2+x^3+x^4+\cdots)+\cdots \\ &=\frac{x}{1-x}+\frac{x^2}{1-x}+\cdots \\&=\frac{1}{1-x}(x+x^2+x^3+\cdots)\\&=\frac{1}{1-x}\frac{x}{1-x}\\&=\frac{x^2}{1-x}\end{align}$$

Here $x=\frac{1}{p-1}<1$ So $\sum_{i=1}^N {\frac{i}{p^i}} \leq\sum_{i=1}^{\infty} {\frac{i}{p^i}}=\frac{p}{(p-1)^2}$

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Thank you A.D., neat argument. – Conan Wong Jan 29 '13 at 19:33
@ConanWong : WELCOME – A.D Jan 29 '13 at 19:44

Hint: it doesn't rqwuire $p$ to be prime. Instead, compute the function:

$$f(z)=\sum_{i=1}^\infty iz^i$$

when $|z|<1$, show that $$f(z)=\frac{z}{(1-z)^2}$$

Then your statement is:

$$\sum_{i=1}^N iz^i \leq f(z)$$

where $0<z<1$, and, in particular, $z=\frac{1}{p}$.

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Thank you Thomas, thanks for pointing out that p need not be prime. – Conan Wong Jan 29 '13 at 19:34

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