Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

A sum geometric question

edit - sorry about the title

Hello, I would appreciate any help with this. I feel like I am overlooking something very obvious, hopefully someone can point it out quickly for me.

SUM [ (1/2^n)(ln (2^n)) ]

The infinite sum of this is apparently 2 ln 2. I cant figure out why it is that though.

Thanks in advance.

(edited 13 years ago)

\displaystyle \sum_{n\ge 0}\frac{\log(2^n)}{2^n} = \sum_{n \ge 0}\frac{n\ln(2)}{2^n} = \ln(2)\sum_{n \ge 0}\frac{n}{2^n}.

Is that what you were over looking -- is it enough of a hint?

OP

Original post by Piecewise

\sum_{n \ge 0}\frac{n}{2^n}.

I dont understand how to show this reduces to 2.

\begin{aligned} \displaystyle \sum_{n \ge 0}\frac{n}{2^n} =\sum_{n \ge 0}n{x}^n\bigg|_{x = \frac{1}{2}} = x\sum_{n \ge 0}nx^{n-1}\bigg|_{x = \frac{1}{2}} = x\left(\sum_{n\ge 0}x^n\right)'\bigg|_{x = \frac{1}{2}} = x\left(\frac{1}{1-x}\right)'\bigg|_{x = \frac{1}{2}} = \frac{x}{(1-x)^2}\bigg|_{x= \frac{1}{2}} = 2. \end{aligned}

(edited 13 years ago)

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