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Hi i have been trying to prove this summation from i=1 to n is less than $\frac{1}{(1-r)^3}$. The only method i could think of is by solving with a geometric series.But i am unsure of the proofing . Please advice ..

$$\sum_{i=1}^n (1+2+3+...+i)r^i < \frac{1}{(1-r)^3}$$ for all $n\ge1$ and $0\lt r\lt1$

using the fact that $$\sum_{i=1}^n ir^i < \frac{r}{(1-r)^2}$$ for all $n\ge1$ and $0\lt r\lt1$

3 Answers 3

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Assume $|x|<1$. If you start with $$ \frac{1}{1-x}=\sum_{n\geq 0} x^n\tag{1} $$ you also have $$ \frac{1}{1-x}\sum_{n\geq 0} a_n x^n = \sum_{n\geq 0}(a_0+\ldots+a_n) x^n\tag{2}$$ by Cauchy product/convolution. So $(1)$ implies $$ \sum_{n\geq 1} n x^n = \frac{x}{(1-x)^2}\tag{3} $$ and by multiplying by $\frac{1}{1-x}$ both sides once again: $$ \sum_{n\geq 1}(1+\ldots+n)x^n = \frac{x}{(1-x)^3}.\tag{4} $$

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Get the closed form for the sum. Then use the derivative for $r^{i+1}$ twice and interchange summations and derivative function.

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    correct me if i am wrong i have the close form for the sum as $$\sum_{i=1}^n ir^i =\sum_{i=1}^ni\sum_{i=1}^nr^i= \frac{(n)(n+1)}{2}*\frac{1-r^{n+1}}{1-r}$$ i am unsure if this is vaild.. – 2017-02-08
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    This is incorrect. The closed form for the sum inside is $\frac{i(i+1)}{2}$, so you have to deal with $i(i+1)$. Also you can't split the sum like this. If you have $(i+1)i r^i = \frac{1}{r^2}\frac{d^2}{dr^2} (r^{i+1})$ – 2017-02-08
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$\displaystyle\sum\limits_{i=1}^n (1+2+3+...+i)r^i < \frac{1}{(1-r)^3}$

This is valid, if

$\displaystyle\sum\limits_{i=1}^n i(i+1)r^i < \frac{2}{(1-r)^3}$

$\displaystyle\sum\limits_{i=1}^n i(i+1)r^i - \sum\limits_{i=1}^n i(i+1)r^{i+1} < \frac{2}{(1-r)^2}$

$\displaystyle -n(n+1)r^{n+1}+2\sum\limits_{i=1}^n ir^i < \frac{2}{(1-r)^2}$

With

$\displaystyle \sum\limits_{i=1}^n ir^i < \frac{r}{(1-r)^2}\enspace $ for $\enspace 0

it's true that

$\displaystyle \sum\limits_{i=1}^n ir^i < \frac{1}{(1-r)^2}+n(n+1)r^{n+1}$

which is equal to the problem and therefore solves it.