Telescoping series: $\sum i^2 x^i$ for $0 < x < 1$ It is asked to find the sum
$$ \sum_{i=1}^{\infty} i^2 x^i $$
Using the telescoping property. 
But I could not find a sequence to write my $s_n$ in function of and apply this.. Does anyone have a hint?
Thanks in advance.
 A: A natural way to use the telescoping property in this context is to start from the identity $$(1-x)\sum_{i\geqslant0}i^2x^i=\sum_{i\geqslant0}i^2x^i-\sum_{i\geqslant1}(i-1)^2x^i=\sum_{i\geqslant1}(2i-1)x^i,$$ that is, $$(1-x)\cdot s_2(x)=2s_1(x)-s_0(x)+1,$$ where, for every $k$ (and with the convention that $0^0=1$), $$s_k(x)=\sum_{i\geqslant0}i^kx^i.$$ A second time, still by the telescoping property, $$(1-x)\cdot s_1(x)=(1-x)\sum_{i\geqslant0}ix^i=\sum_{i\geqslant1}ix^i-\sum_{i\geqslant1}(i-1)x^i=\sum_{i\geqslant1}x^i=s_0(x)-1.$$ And a third time, still by the telescoping property, $$(1-x)\cdot s_0(x)=(1-x)\sum_{i\geqslant0}x^i=\sum_{i\geqslant0}x^i-\sum_{i\geqslant1}x^i=1.$$ All this leads to $$(1-x)^2\cdot s_1(x)=(1-x)\cdot s_0(x)-(1-x)=1-(1-x)=x,$$
which implies
$$(1-x)^3\cdot s_2(x)=2(1-x)^2s_1(x)-(1-x)^2s_0(x)+(1-x)^2=2x-(1-x)+(1-x)^2=x+x^2,$$ that is, $$s_2(x)=\frac{x(1+x)}{(1-x)^3}.$$
A: Here's a way to sum the series.  It doesn't mention the telescoping property and it has a gap in the logic, namely that the derivative of the sum is sometimes not the sum of the derivatives when the number of terms is infinite.  I may come back and address those questions later.
\begin{align}
\sum_{i=1}^\infty i^2 x^i & = x^2\sum_{i=1}^\infty i(i-1)x^{i-2} + x\sum_{i=1}^\infty ix^{i-1} \\[10pt]
& = x^2 \frac{d^2}{dx^2}\sum_{i=2}^\infty x^i + x\frac{d}{dx}\sum_{i=1}^\infty x^i.
\end{align}
Notice that the first sum now starts at $i=1$ and that there is a reason for that.
The remaining series are geometric: each has a first term and a common ratio.  After summing them, do the differentions.
A: When the telescoping techniques is used to evaluate the sum of a series, firstly the terms in the sum are possibly rewritten themselves as two or more terms. Then with the newer expanded terms, it is always shown that $$S_n=\left(\mbox{a fixed number of initial terms}\right)+\left(\mbox{many terms}\right)+\left(\mbox{a fixed number of final terms}\right)$$
where the middle terms sum to zero, and the final terms approach zero. So if this is your task, then you are expected to rewrite a finite number of initial terms in a way that they sum to the actual sum, which is $$\frac{x+x^2}{(1-x)^3}$$ and at that point it makes one wonder if the instructions to use the telescoping property are simply a mistake.
