# Thread: SUM of N / 2^N

1. ## SUM of N / 2^N

Hello:

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

Thanks!

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

3. 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.

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

5. 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.

6. 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.

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