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Calculate: $\dfrac{1}{2}+\dfrac{2}{2^²}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...=?$

I try

$\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...\dfrac{1}{2^n}+\dfrac{1}{2^{n+1}}+\dfrac{1}{2^{ n+2}}$

$\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}+\dfrac{\dfrac{1}{4}}{1-\dfrac{1}{2}}+\dfrac{\dfrac{1}{8}}{1-\dfrac{1}{2}}+ ...\dfrac{\dfrac{1}{2^n}}{1-\dfrac{1}{2}}\\ 1+\dfrac{1}{2}+\dfrac{1}{4}+...\dfrac{2}{2^n}$

how to finish

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  • $\begingroup$ Your sum can be rewritten as $\sum_{n=1}^\infty nx^n$ where $x=1/2$. Does this help? $\endgroup$
    – Clayton
    Commented May 12, 2023 at 12:09
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    $\begingroup$ You are only one step from solving it yourselves, and I like that solution! The last sum is just $1+1/2+1/4+1/8+\ldots = 2$. $\endgroup$
    – user700480
    Commented May 12, 2023 at 12:30

2 Answers 2

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Let $x=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\ldots$

Then, $2x=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+\ldots$

Then, $2x-1-x=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\ldots=1$

So, $2x-1-x=1$ i.e. $x=2$.

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It is

$$ \sum_n \dfrac{n}{2^n} $$

So take

$$ \sum_n x^n=\dfrac{1}{1-x} $$

Derivating

$$ \sum_n nx^{n-1}=\dfrac{1}{(1-x)^2} $$

$x=1/2$

$$ \sum_n n(\dfrac{1}{2})^{n-1} $$

Adjust and you get the result.

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  • $\begingroup$ The OP may not know derivatives. $\endgroup$
    – user700480
    Commented May 12, 2023 at 12:25
  • $\begingroup$ What is the OP ? $\endgroup$
    – EDX
    Commented May 12, 2023 at 12:27
  • $\begingroup$ Original Poster $\endgroup$
    – user700480
    Commented May 12, 2023 at 12:27

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