Hello:

I wonder how do you prove that the SUM of N / 2^N (N=1 to infinity) is N / (N-1)^2.

Thanks!

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- Oct 24th 2007, 12:32 PM #1

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- Oct 24th 2007, 12:48 PM #2

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Hello, maltz!

Please check the original wording of the problem.

. . As written, it makes no sense.

Prove: .$\displaystyle \sum^{\infty}_{n=1}\frac{n}{2^n} \;=\;\frac{n}{(n-1)^2} $

If $\displaystyle n \to \infty$, there would be no $\displaystyle n$ on the right side.

If you meant: .$\displaystyle \sum^n_{k=1}\frac{k}{2^k} \;=\;\frac{n}{(n-1)^2} $, .that makes more sense,

. . . . but unfortunately, it is not true.

- Oct 24th 2007, 01:18 PM #6

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- Oct 24th 2007, 01:34 PM #7
$\displaystyle \sum\limits_{N = 0}^\infty {x^N } = \frac{1}{{1 - x}},\;\left| x \right| < 1$

Differentiate: $\displaystyle \sum\limits_{N = 1}^\infty {Nx^{N - 1} } = \frac{1}{{\left( {1 - x} \right)^2 }}$.

Multiply by x: $\displaystyle \sum\limits_{N = 1}^\infty {Nx^N } = \frac{x}{{\left( {1 - x} \right)^2 }}$.

Let $\displaystyle x = \frac{1}{K},\;K > 1$