SUM of N / 2^N

**maltz** · October 24th 2007, 12:32 PM

Hello:

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

Thanks!

**Plato** · October 24th 2007, 12:48 PM

Originally Posted by maltz

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

One does not because $\sum\limits_{N = 1}^\infty {\frac{N}{{2^N }}} = \frac{{\left( {\frac{1}{2}} \right)}}{{\left( {1 - \frac{1}{2}} \right)^2 }} = 2$ .

**maltz** · October 24th 2007, 01:01 PM

Ops I am sorry. I wrote the question wrong.
It should be SUM (N / K^N) (N = 1 to infinity, K is a positive integer),
and the result is N/(K-1)^2

What is the formula? Thanks again.

**Plato** · October 24th 2007, 01:12 PM

Once again you have written a false statement.
This is true:
$\sum\limits_{N = 1}^\infty {\frac{N}{{K^N }}} = \frac{{\left( {\frac{1}{K}} \right)}}{{\left( {1 - \frac{1}{K}} \right)^2 }} = \frac{K}{{\left( {K - 1} \right)^2 }},\;K > 1.$

**Soroban** · October 24th 2007, 01:13 PM

Hello, maltz!

Please check the original wording of the problem.
. . As written, it makes no sense.

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

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

If you meant: . $\sum^n_{k=1}\frac{k}{2^k} \;=\;\frac{n}{(n-1)^2}$ , .that makes more sense,
. . . . but unfortunately, it is not true.

**maltz** · October 24th 2007, 01:18 PM

Yeah I suck... sorry.
So it is SUM (N/K^N) = K/(K-1)^2
I am just wondering how you can proof it instead of memorizing a formula.
That would be greatly appreciated.

**Plato** · October 24th 2007, 01:34 PM

$\sum\limits_{N = 0}^\infty {x^N } = \frac{1}{{1 - x}},\;\left| x \right| < 1$

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

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

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

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