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This question already has an answer here:

$$\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}+\dfrac{5}{64}+\cdots\infty$$

This summation was irritating me from the start,I don't know how to attempt this ,tried unsuccessful attempts though.

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marked as duplicate by lab bhattacharjee, Daniel Fischer May 6 '17 at 12:40

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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$$\begin{align} S&=\qquad \frac 14+\frac 28+\frac 3{16}+\frac 4{32}+\frac 5{64}+\cdots\tag{1}\\ 2S&=\frac 12+\frac 24+\frac 38+\frac 4{16}+\frac 5{32}\cdots\tag{2}\\ (2)-(1):\qquad\\ S&=\frac 12+\frac 14+\frac 18+\frac 1{16}+\frac 1{32}\cdots\\ &=\frac {\frac 12}{1-\frac 12}\\ &=\color{red}1 \end{align}$$

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The series is $$ \frac{1}{4}\sum_{n\ge1}nx^{n-1} $$ where $x=1/2$. Does this remind you about something?

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One may start with the standard evaluation, $$ 1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1 $$ Then by differentiating $(1)$ one gets $$ 1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}-\frac{(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2 $$ by multiplying by $x^2$ and by making $n \to +\infty$ in $(2)$, using $|x|<1$, one has

$$ \sum_{n=1}^\infty nx^{n+1}=\frac{x^2}{(1-x)^2}. \tag3 $$

Then one may put $x=\dfrac12$ to obtain an answer to the given sum.

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Apply $(1-x)^{-2}=1+2x+3x^2+4x^3+\cdots\infty$.

Now,

$\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}+\dfrac{5}{64}+\cdots\infty\\ =\dfrac{1}{4}\left(1+\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+\cdots\infty\right)\\ =\dfrac{1}{4}\left(1-\dfrac{1}{2}\right)^{-2}\hspace{25pt}\text{ here }x=\dfrac{1}{2}.\\ =\dfrac{1}{4}\times4=1.$

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