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Differentiate both sides of the geometric sum with respect to $r$.$$\sum_{i=0}^n r^i = \frac{1-r^{n+1}}{1-r}$$ Use the result to show that $$\sum_{i=1}^n ir^i < \frac{r}{(1-r)^2} \text{ for all } n\ge1 .$$

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    Uh...and what have you tried...? For starters, what did you get after differentiation? – 2017-02-06
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    If you know that $(1)$ $\dfrac d {dr} r^i = ir^{i-1}$ and $(2)$ the quotient rule and $(3)$ the derivative of the sum of several terms is the sum of their derivatives, then you've got it. – 2017-02-06
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    Maybe of interest http://math.stackexchange.com/questions/1652794/find-a-closed-formula-for-sum-n-1-infty-nxn-1 – 2017-02-07

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