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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

peta arantes
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  • Your sum can be rewritten as $\sum_{n=1}^\infty nx^n$ where $x=1/2$. Does this help? – Clayton May 12 '23 at 12:09
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    You are only one step from solving it yourselves, and I like that solution! The last sum is just $1+1/2+1/4+1/8+\ldots = 2$. –  May 12 '23 at 12:30

2 Answers2

2

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.

EDX
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2

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. $x=2$.