0
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Your sum can be rewritten as $\sum_{n=1}^\infty nx^n$ where $x=1/2$. Does this help? $\endgroup$
    – Clayton
    May 12 at 12:09
  • 2
    $\begingroup$ 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
    May 12 at 12:30

2 Answers 2

Reset to default
2
$\begingroup$

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. $x=2$.

$\endgroup$
0
2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ The OP may not know derivatives. $\endgroup$
    – user700480
    May 12 at 12:25
  • $\begingroup$ What is the OP ? $\endgroup$
    – EDX
    May 12 at 12:27
  • $\begingroup$ Original Poster $\endgroup$
    – user700480
    May 12 at 12:27

Not the answer you're looking for? Browse other questions tagged .