Induction Question Summation and Inequality I'm trying to prove that 
$$\sum_{i=1}^n (1+2+3+...+i)r^i < \frac{1}{(1-r)^3}\,\,\,\,\,\,\,for\,all\,n\ge1 \,and\,0\lt r\lt1.$$ 
Using the fact that
$$\sum_{i=1}^n ir^i < \frac{1}{(1-r)^2}\,\,\,\,\,\,\,for\,all\,n\ge1 \,and\,0\lt r\lt1.$$
 A: First, note that we can write the sum of interest as
$$\sum_{i=1}^n \color{blue}{\left(1+2+3+\cdots +i\right)}r^i=\sum_{i=1}^n \color{blue}{\sum_{j=1}^i(j) }r^i \tag 1$$

Second, interchanging the order of summation in $(1)$ yields
$$\sum_{i=1}^n \left(1+2+3+\cdots +i\right)r^i=\sum_{j=1}^n j \left(\sum_{i=j}^n r^i \right)\tag 2$$

Third, summing the geometric progression in $(2)$, we obtain
$$\sum_{i=1}^n \left(1+2+3+\cdots +i\right)r^i=\sum_{j=1}^n j \left(\frac{r^j-r^{n+1}}{1-r}\right)\tag 3$$

Finally, using $r^j-r^{n+1}<\,r^j$ along with the given inequality $\sum_{j=1}^n jr^j<\frac{1}{(1-r)^2}$ in $(3)$ reveals
$$\sum_{i=1}^n \left(1+2+3+\cdots +i\right)r^i<\frac{1}{(1-r)^3}$$
as was to be shown!
A: Observe
\begin{align}
\sum^n_{i=1} (1+2+\ldots+i) r^i = \sum^n_{i=1}\frac{i(i+1)}{2}r^i = \frac{1}{2}\sum^n_{i=1}i^2r^i + \frac{1}{2}\sum^n_{i=1}ir^i.
\end{align}
Next, observe
\begin{align}
\sum^n_{i=1}i^2r^i <\sum^\infty_{i=1}i^2r^i = \frac{r(1+r)}{(1-r)^3}<\frac{1+r}{(1-r)^3}
\end{align}
which means
\begin{align}
\sum^n_{i=1} (1+2+\ldots+i) r^i <\frac{1+r}{2(1-r)^3}+\frac{1-r}{2(1-r)^3} = \frac{1}{(1-r)^3}.
\end{align}
Edit: This is probably not the best way to solve the problem because it involves using
\begin{align}
\sum^\infty_{i=1}i^2r^i= \frac{r(1+r)}{(1-r)^3}
\end{align}
which kind of destroys the spirit of the problem. 
Edit: Let us observe
\begin{align}
\sum^\infty_{i=1}i^2r^i =&\ r\sum^\infty_{i=1}i^2r^{i-1} = r\frac{d}{dr}\sum^\infty_{i=1}ir^i
=r\frac{d}{dr}\left(r\sum^\infty_{i=1}ir^{i-1} \right)\\
=&\ r\frac{d}{dr}\left(\frac{r}{(1-r)^2} \right)= r\frac{(1-r)^2+2r(1-r)}{(1-r)^4}\\
=&\ r\frac{(1-r)+2r}{(1-r)^3} = \frac{r(1+r)}{(1-r)^3}.
\end{align}
