Evaluation of a power series: looking for a suitable function

Question

I'd like to find the sum of the series $\sum_{n=0}^{\infty} \frac{2n^2+3}{2 \cdot 3^n}$. I thought that, putting $x$ instead of $3$, gives a power series, so my hope is that it would be a power series of a known function, but I got stuck there. I also compute that the ratio of convergence of such power series is $1$.I feel I am missing something, can you help?

Answer 1

You have the right idea, but I found it simpler to first break up the series into two smaller ones, i.e. $\sum_{n=0}^{\infty}\frac{2n^2+3}{2\cdot3^n}=\sum_{n=0}^{\infty}\frac{2n^2}{2\cdot3^n}+\sum_{n=0}^{\infty}\frac{3}{2\cdot3^n}$. Simplifying these expressions slightly leads to $\sum_{n=0}^{\infty}\frac{2n^2+3}{2\cdot3^n}=\sum_{n=0}^{\infty}\frac{n^2}{3^n}+\sum_{n=0}^{\infty}\frac{1}{2\cdot3^{n-1}}$

Let us focus on the rightmost series first, since it's the simplest. The first term is $n=0$, and when $n=0$, $\frac{1}{2\cdot3^{n-1}}=\frac{1}{2\cdot3^{-1}}=\frac{1}{\frac{2}{3}}=\frac{3}{2}$. The following terms are obtained by multiplying the previous one by $\frac{1}{3}$. We thus identify this as a geometric series with initial term $\frac{3}{2}$ and common ratio $\frac{1}{3}$; as such, its sum is $\frac{\frac{3}{2}}{1-\frac{1}{3}}=\frac{\frac{3}{2}}{\frac{2}{3}}=\frac{9}{4}$.

Now for the first series on the RHS. Define $f(x)=\sum_{n=0}^{\infty}n^2x^n$; if $\frac{1}{3}$ is substituted for $x$ then we obtain precisely the series we want the sum of, i.e. $f(\frac{1}{3})=\sum_{n=0}^{\infty}\frac{n^2}{3^n}$. But how do we compute this sum? The answer is, again, the geometric series. But first, observe the first few terms: $\sum_{n=0}^{\infty}\frac{n^2}{3^n}=0\cdot3^0+1\cdot3^{-1}+4\cdot3^{-2}+9\cdot3^{-3}+...$

The point of that exercise will become clear in just a little bit, but for now, consider the series $\sum_{n=0}^{\infty}x^n$. As you know this is the geometric series with the very simple expansion of $x^0+x^1+x^2+x^3+...$ and the closed form of $\frac{1}{1-x}$.

What happens if we take the derivative with respect to $x$ of each of these representations? Setting the expansion and the closed form equal to each other, taking the derivative of the expansion yields:

$\frac{d}{dx}[x^0+x^1+x^2+x^3+x^4+...]=1x^0+2x^1+3x^2+4x^3...$

While taking the derivative of the closed form yields $\frac{1}{(1-x)^2}$.

These things are the same object, and so we might as well multiply by $x$. Doing this leads to the following:

$x\frac{d}{dx}[x^0+x^1+x^2+x^3+...]=x\frac{d}{dx}[\frac{1}{1-x}]$

$\Rightarrow 1x^1+2x^2+3x^3+4x^4+... =\frac{x}{(1-x)^2}$

Now the really cool thing about this is that these are both equal to $\sum_{n=0}^{\infty}nx^n$! Which is almost what you want, but not quite. Can you guess how we get to the series we want? By taking the derivative and multiplying by $x$ again, of course! Taking the derivative of the expansion:

$\frac{d}{dx}[1x^1+2x^2+3x^3+4x^4+...]=1x^0+4x^1+9x^2+16x^3+...$

Taking the derivative of the closed form:

$\frac{d}{dx}[\frac{x}{(1-x)^2}]=\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\frac{(1-x)((1-x)+2x)}{(1-x)^4}=\frac{(1-x)(1+x)}{(1-x)^4}=\frac{1-x^2}{(1-x)^4}$

Now multiplying the RHS of the first equation and the RHS of the second equation by $x$ yields the following third equation:

$1x^1+4x^2+9x^3+16x^4+...=\frac{x-x^3}{(1-x)^4}$

We can see that we now have a closed form for $f(x)$, which is precisely what we wanted. And since $\sum_{n=0}^{\infty}\frac{n^2}{3^n}=f(\frac{1}{3})$, all we need to do now is to plug in $\frac{1}{3}$ for $x$ in our closed-form formula:

$\sum_{n=0}^{\infty}\frac{n^2}{3^n}=f(\frac{1}{3})=\frac{\frac{1}{3}-\frac{1}{3^3}}{(1-\frac{1}{3})^4}=\frac{\frac{8}{27}}{\frac{16}{81}}=\frac{8\cdot81}{16\cdot27}=\frac{2^33^4}{2^43^3}=\frac{3}{2}$

Almost there! Recall that the sum of the original series is equal to the sum of the sums of our two subseries; i.e. $\sum_{n=0}^{\infty}\frac{2n^2+3}{2\cdot3^n}=\frac{9}{4}+\frac{3}{2}=\frac{9+2\cdot3}{4}=\frac{15}{4}=3.75$

And that's it!

Caveat: the method used to find the trickier sum only worked because that series was convergent, which I didn't prove. In general, you can only play around with series like this when they converge. For more information, I highly recommend checking out Herbert Wilf's publicly available (and free!) book, generatingfunctionology. The technique I used came directly from there, and much more besides. It won't steer you wrong and is very digestible!

Addendum: just to note, you can use this technique to find the sum of any series of the form $\sum_{n=0}^{\infty}\frac{an^k+b}{ab^n}$, and indeed any series — provided they converge — of the form $\sum_{n=0}^{\infty}\frac{an^k+b}{cd^n}$ (you would split the sum as before and factor out $\frac{a}{c}$ and $\frac{b}{c}$ as constants in front of the summation sign, and proceed as usual)