Prove that $\sum_{i=1}^n (1+2+3+...+i)r^i < \frac{1}{(1-r)^3}$ for all $n\ge1$ and $0\lt r\lt1$ 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 $0\lt r\lt1$
 A: 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:
$$ \sum_{n\geq 1}(1+\ldots+n)x^n = \frac{x}{(1-x)^3}.\tag{4} $$
A: Get the closed form for the sum. Then use the derivative for $r^{i+1}$ twice and interchange summations and derivative function.
A: $\displaystyle\sum\limits_{i=1}^n (1+2+3+...+i)r^i < \frac{1}{(1-r)^3}$ 
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 $\enspace 0<r<1$ 
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. 
