<s>[INST] Differentiation method for evaluating $ \sum_{n=1}^\infty \frac{n^2}{3^n} $ I evaluated the following infinite sum (the original and broader question regarding this sum can be found at Evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n} $).
$$ \sum_{n=1}^\infty \frac{n^2}{3^n} $$
However, I'm getting the feeling that I made some mistake(s) during my evaluations. My particular concern regards the changes made in the sum's index. The following is what I did.
I first defined a power series as the function f.
$$ \sum_{n=1}^\infty \frac{n^2}{3^n} = \sum_{n=1}^\infty n^2 (\frac{1}{3})^n = f(\frac{1}{3}) \Rightarrow f(x) = \sum_{n=1}^\infty n^2 x^n $$
I then attempted to manipulate the sum in order to transform it through the geometric series. This is where I'm quite unsure whether I did everything correctly. One of the things I did here looks wrong to me, but somehow, I still ended up with the correct answer (which is 3/2).
$$
\begin{align*}
f(x) &= \sum_{n=1}^\infty n^2 x^n
\\ &= x \sum_{n=1}^\infty n^2 x^{n-1}
\\ &= x \frac{d}{dx} ( \sum_{n=0}^\infty n x^n )
\\ &= x \frac{d}{dx} ( x \sum_{n=0}^\infty n x^{n-1} )
\\ &= x \frac{d}{dx} ( x \frac{d}{dx} ( \sum_{n=-1}^\infty x^n ) )
\\ &= x \frac{d}{dx} ( x \frac{d}{dx} ( \frac{1}{1-x} ) )
\\ &= x \frac{d}{dx} ( x \frac{1}{(1-x)^2} )
\\ &= x \frac{d}{dx} ( \frac{x}{(1-x)^2} )
\\ &= x \frac{1+x}{(1-x)^3}
\\ &= \frac{x(1+x)}{(1-x)^3}
\\ &\Rightarrow f(\frac{1}{3}) = \frac{3}{2}
\end{align*}
$$
The main concern of mine is the transition step to the closed-form geometric series. Of course, the proper equation for a geometric series is this:
$$ \sum_{n=0}^\infty x^n = \frac{1}{1-x} $$
However, what I did is this:
$$ \sum_{n=-1}^\infty x^n = \frac{1}{1-x} $$
The difference here is that the starting index is -1 instead of 0. This makes my translation of the sum incorrect. And yet, I still get the correct answer. On the other hand, I've tried the correct(?) form of an infinite geometric series that starts at n=-1:
$$ \sum_{n=-1}^\infty x^n = \frac{\frac{1}{x}}{1-x} = \frac{1}{x(1-x)} $$
However, this yields an incorrect answer.
I'm guessing that the final index I should've had for the sum was n=0 instead of n=-1. I'm guessing I did something wrong with the index shifts caused by the derivatives? Either way, I'm not seeing it.
I do realize that there are other ways to go about evaluating this sum, but I'd really like to understand this derivative method.
 [/INST] You could introduce a step in which you write
$$
\begin{align}
f(x) & = \cdots \\
     & = x \frac{d}{dx} \left( x \sum_{n=0}^\infty nx^{n-1} \right) \\
     & = \color{red}{
         x \frac{d}{dx} \left( x \sum_{n=1}^\infty nx^{n-1} \right)} \\
     & = x \frac{d}{dx} \left( x \frac{d}{dx}
                        \left( \sum_{n=0}^\infty x^n \right) \right) \\
     & = \cdots
\end{align}
$$
since $nx^{n-1} = 0$ when $n = 0$.  That should resolve matters, although it isn't in fact necessary, because$\ldots$
ETA: Actually, your line is in error.  You write
$$
\begin{align}
f(x) & = \cdots \\
     & = x \frac{d}{dx} \left( x \sum_{n=0}^\infty nx^{n-1} \right) \\
     & = x \frac{d}{dx} \left( x \frac{d}{dx}
                        \left( \sum_{n=-1}^\infty x^n \right) \right) \\
     & = \cdots
\end{align}
$$
but the index limit should have remained $n = 0$, since
$$
\begin{align}
f(x) & = \cdots \\
     & = x \frac{d}{dx} \left( x \sum_{n=0}^\infty nx^{n-1} \right) \\
     & = \color{red}{
         x \frac{d}{dx} \left( x \sum_{n=0}^\infty \frac{d}{dx} x^n \right)} \\
     & = x \frac{d}{dx} \left( x \frac{d}{dx}
                        \left( \sum_{n=0}^\infty x^n \right) \right) \\
     & = \cdots
\end{align}
$$
</s>