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I am trying to find find the value of $\sum_{n=2}^{\infty}n{\frac1{2^{n-1}}}$

I started by substituting m=n-1 to get $\sum_{m=1}^{\infty}(m+1){\frac1{2^{m}}}$ but I am not sure where to go from here.

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Hint:

$$S:=\frac22+\frac34+\frac48+\frac5{16}+\cdots$$

and

$$S=\frac{1+1}2+\frac{2+1}4+\frac{3+1}8+\frac{4+1}{16}+\cdots\\ =\frac12+\frac12\left(\frac22+\frac34+\frac48+\frac5{16}+\cdots\right)+\left(\frac12+\frac14+\frac18+\frac1{16}+\cdots\right)$$

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    $\begingroup$ I like this answer the best out of the two. It helped me see the convergent series then I can use the formula for convergent geometric series to find the values of the infinite sums. Thanks! $\endgroup$ – mattsprestige Oct 1 '19 at 0:54
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HINT

Use that by derivative

$$\sum x^n=f(x) \implies \sum nx^{n-1}=f’(x) $$

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