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i need some ideas to solve $$\sum_{n=1}^\infty n\cdot\left(\frac12\right)^n$$ I prove that the series converges to using ratio method, but i dont know how to find the sum.

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marked as duplicate by Guy Fsone, Namaste, Jack, Hans Lundmark, user223391 Jan 15 '18 at 18:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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$$S=\sum_{n=1}^\infty n\cdot\left(\frac12\right)^n$$ Then $$\begin{align}S&=\frac12+2\cdot\frac14+3\cdot\frac18+4\cdot\frac1{16}+\cdots\\ &=\left(\frac12+\frac14+\frac18+\cdots\right)+\left(\frac14+\frac18+\cdots\right)+\left(\frac18+\cdots\right)+\cdots\\ &=\left(\frac12+\frac14+\frac18+\cdots\right)\cdot\left(1+\frac12+\frac14+\cdots\right)\\ &=1\cdot 2\\ &=2\end{align}$$

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Start with

$$\sum_{n=0}^\infty x^{n} = \frac{1}{1-x}.$$

Then take the derivative of both sides

$$\sum_{n=1}^\infty nx^{n-1} = \frac{1}{(1-x)^2}.$$

Multiply by $x$

$$\sum_{n=1}^\infty nx^{n} = \frac{x}{(1-x)^2}.$$

Then plug in $x=1/2.$

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use that for the finite sum is $$\sum_{i=1}^n\frac{i}{2^i}=-2\, \left( 1/2 \right) ^{n+1} \left( n+1 \right) -2\, \left( 1/2 \right) ^{n+1}+2 $$ compute the Limit for $n$ tends to infinity

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$$\sum_{n=1}^{\infty}\frac{n}{2^{n}}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{n}{2^{n-1}}=\frac{1}{2}\left(1+x+x^2+...\right)'_{x=\frac{1}{2}}=\frac{1}{2}\left(\frac{1}{1-x}\right)'_{x=\frac{1}{2}}=$$ $$=\frac{1}{2}\cdot\frac{1}{\left(1-\frac{1}{2}\right)^2}=2$$

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Are you familiar with the usual method for summing a finite geometric series?

Let $\displaystyle S_n = \sum_{k=1}^n k\cdot\left(\frac12\right)^k= \left(1 \cdot \dfrac 12 + 2 \cdot \dfrac {1}{2^2}+ 3 \cdot \dfrac {1}{2^3} + \cdots + n \cdot \dfrac {1}{2^n} \right)$

Then $\displaystyle \dfrac 12 S_n= \sum_{k=1}^n k\cdot\left(\frac12\right)^{k+1} = \left(1 \cdot \dfrac {1}{2^2} + 2 \cdot \dfrac {1}{2^3}+ 3 \cdot \dfrac {1}{2^4} + \cdots + (n-1) \cdot \dfrac {1}{2^n} + n \cdot \dfrac {1}{2^{n+1}} \right)$

So

$$S_n - \dfrac 12 S_n= \dfrac 12 + \dfrac {1}{2^2} + \dfrac {1}{2^3} + \cdots + \dfrac {1}{2^n} + n \cdot \dfrac {1}{2^{n+1}} $$

$$=\dfrac {1 - (\frac {1}{2})^{n+1}}{1 - \frac 12}-1 + n \cdot \dfrac {1}{2^{n+1}} $$

Thus

$$\dfrac 12 S_n= 2-\left( \dfrac{1}{2^n} \right)-1 - n \cdot \dfrac {1}{2^{n+1}}$$

$$S_n=4-\left( \dfrac{1}{2^{n-1}} \right)-2 - n \cdot \dfrac {1}{2^{n}}$$

So

$$\sum_{k=1}^\infty k\cdot\left(\frac12\right)^k = \displaystyle \lim_{n \to \infty} S_n =2$$

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Other technique different than the showed in other answers make use of summation by parts together with particular algebraic rules named generally finite calculus.

In this fashion we want to consider the indefinite sum $\sum k x^k\,\delta k$. Using the techniques described in the last linked PDF we find that

$$\sum k x^k\,\delta k=\frac{x^k}{x-1}\cdot k-\frac1{x-1}\sum x^{k+1}\,\delta k\\\implies\sum_{k=0}^\infty kx^k=\left[\frac{x^k}{x-1}\cdot k\right]_{k=0}^{k\to\infty}-\frac1{x-1}\sum_{k=0}^\infty x^{k+1}$$

Choosing $x=1/2$ we find that

$$\sum_{k=0}^\infty k\left(\frac12\right)^k=\lim_{k\to\infty}\frac{(1/2)^k\cdot k}{-1/2}+2\sum_{k=1}^\infty (1/2)^k=0+2=2$$

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