2
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$$\sum_{k=1}^{\infty} \frac{k}{2^k}$$

I know that $\sum_{k=1}^{\infty} \frac{1}{2^k}=1$ so the answer is $\frac{k(k+1)}{2}*1$?

the answer is 2

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  • $\begingroup$ The result could not depend on $k$! Have you tried wolfram alfa? $\endgroup$
    – iiivooo
    Mar 20, 2016 at 17:14
  • $\begingroup$ You cannot multiply them together like that $\endgroup$
    – Shuri2060
    Mar 20, 2016 at 17:14
  • 1
    $\begingroup$ Also, $$\sum_{k=1}^{\infty} k \neq \frac{k(k+1)}{2}$$ $\endgroup$ Mar 20, 2016 at 17:15

2 Answers 2

3
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Hint: for $|x|<1$, $$\sum \limits_{k=1}^{\infty} x^n=\frac{1}{1-x}. $$

Try differentiating both sides of this equation and let $x= \frac{1}{2}$.

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Here is a useful finite evaluation: $$ 1+r+r^2+...+r^n=\frac{1-r^{n+1}}{1-r}, \quad |r|<1. \tag1 $$ Then by differentiating $(1)$ you get $$ 1+2r+3r^2+...+nr^{n-1}=\frac{1-r^{n+1}}{(1-r)^2}+\frac{-(n+1)r^{n}}{1-r}, \quad |r|<1, \tag2 $$ and by making $n \to +\infty$ in $(2)$, using $|r|<1$, gives

$$ 1+2r+3r^2+...+nr^{n-1}+...=\frac{1}{(1-r)^2} \tag3 $$

If you multiply $(3)$ by $r$ and set $r=\dfrac12$, you obtain an answer to your question.

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