Solution of sum

Question

What is an easy way to find the solution of this sum?

$$\displaystyle\sum_{i=1}^{\infty} i3\Bigg(\frac{1}{3}\Bigg)^{i+1}$$

wolframalpha solution: $\frac{3}{4}$

Roll a pair of dice, one after the other, and put the result of the first toss of a die in the numerator, and the result of the second toss of the die in the denominator. Looks like WA rolled first a three, then a four. — amWhy, Mar 07 '18 at 00:04
this serie is completely weird $\displaystyle\sum_{i=1}^{\infty} i3\Bigg(\frac{1}{3}\Bigg)^{i+1}=\displaystyle\sum_{i=1}^{\infty} i\Bigg(\frac{1}{3}\Bigg)^{i}$ — user577215664, Mar 07 '18 at 00:15

score 0 · Accepted Answer · answered Mar 07 '18 at 00:05

0

Somewhat more generally, if $|r| < 1$ $$\sum_{i=1}^\infty a i r^i = r \frac{d}{dr} \sum_{i=1}^\infty a r^i$$ Now use the formula for geometric series.

answered Mar 07 '18 at 00:05

Robert Israel

448,999

but starting from $i=1$ – G Cab Mar 07 '18 at 00:11
Starting from either $1$ or $0$: $i=0$ doesn't contribute. – Robert Israel Mar 07 '18 at 00:31

Solution of sum

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