Proof that $\sum_{j=1}^{\infty} \frac{\theta^j}{j!} \cdot e^{-\theta} \cdot j^2$ reduces to $(\theta^2 + \theta)$

Question

Consider the following sum:

$$\sum_{j=1}^{\infty} \frac{\theta^j}{j!} \cdot e^{-\theta} \cdot j^2$$

This should reduce to:

$$(\theta^2 + \theta)$$

My approach:

$$= e^{-\theta} \cdot \sum_{j=1}^{\infty} \frac{\theta^j}{j!} \cdot j^2$$ $$= e^{-\theta} \cdot \sum_{j=1}^{\infty} \frac{\theta^j}{(j-1)!} \cdot j$$ $$= e^{-\theta} \cdot \theta \cdot \sum_{j=1}^{\infty} \frac{\theta^{(j-1)}}{(j-1)!} \cdot j$$

Since we have a taylor series, we know that $\sum_{j=1}^{\infty} \frac{\theta^{(j-1)}}{(j-1)!}=e^{\theta}$. But there is still $j$ in this sum. Do you know how to proceed?

Answer 1

Let's proceed right ahead, taking the common exponent outside of the sum. Notice that the trick to the question is just to take out all degrees of $j^n$ with the factorial. To do so we use the identity $j^2 = j(j-1+1)$ and notice the fact that the series looks suspiciously like that of the exponent expansion. I should warn that I'll omit specific indexing on $j$, implying the summation from the first non-zero term. $$ e^{-\theta}\sum_{j=1}^\infty{\frac{{\theta}^{j}}{j!}}\cdot j^2 = e^{-\theta}\sum_{j=1}^\infty{\frac{{\theta}^{j}}{j!}}\cdot j(j-1+1) = e^{-\theta}\sum_{j=1}^\infty{\frac{{\theta}^{j}}{j!}}\cdot (j(j-1)+j) = e^{-\theta} \left[ \sum_{j=1}^\infty{\frac{{\theta}^{j}}{j!}}\cdot j(j-1) + \sum_{j=1}^\infty{\frac{{\theta}^{j}}{j!}}\cdot j \right] = e^{-\theta} \left[ \sum_{j}^\infty{\frac{{\theta}^{j}}{(j-2)!}} + \sum_{j}^\infty{\frac{{\theta}^{j}}{(j-1)!}} \right] $$

Now all that's left is to recognise the expansion for the exponent with a few extra multiples of $\theta$. To make it clearer, let's factor out needed multiples that are independent of $j$ when summing: $$ e^{-\theta} \left[ \sum_{j}^\infty{\frac{{\theta}^{2}\cdot{\theta}^{j-2}}{(j-2)!}} + \sum_{j}^\infty{\frac{{\theta}\cdot{\theta}^{j-1}}{(j-1)!}} \right] = e^{-\theta} \left[ {\theta}^{2}\cdot\sum_{j}^\infty{\frac{{\theta}^{j-2}}{(j-2)!}} + {\theta}\cdot\sum_{j}^\infty{\frac{{\theta}^{j-1}}{(j-1)!}} \right] $$

Reindexing the summation indices as $j-2=k$ and $j-1=k'$ in the first and second series expansions respectively and recognising common exponents, one obtains: $$ e^{-\theta} \left[ {\theta}^{2}\cdot\sum_{k=0}^\infty{\frac{{\theta}^{k}}{k!}} + {\theta}\cdot\sum_{k'=0}^\infty{\frac{{\theta}^{k'}}{k'!}} \right] = e^{-\theta} \cdot \sum_{k=0}^\infty{\frac{{\theta}^{k}}{k!}} \cdot ( {\theta}^{2} + {\theta} )= ( {\theta}^{2} + {\theta} ) \cdot e^{-\theta} \cdot \sum_{k=0}^\infty{\frac{{\theta}^{k}}{k!}} = ({\theta}^{2} + {\theta}) \cdot e^{-\theta} \cdot e^{\theta} = {\theta}^{2} + {\theta} $$

Answer 2

$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$ $$e^x - 1 = \sum_{n=1}^{\infty} \frac{x^n}{n!} $$ $$x\frac{d}{dx}x\frac{d}{dx}(e^x - 1) = x\frac{d}{dx}x\frac{d}{dx}\sum_{n=1}^{\infty} \frac{x^n}{n!} = \sum_{n=1}^{\infty} \frac{x^n}{n!} n^2 $$ $$e^{-x}x\frac{d}{dx}x\frac{d}{dx}(e^x - 1) = x\frac{d}{dx}x\frac{d}{dx}\sum_{n=1}^{\infty} \frac{x^n}{n!} = \sum_{n=1}^{\infty} \frac{x^n}{n!} e^{-x} n^2 $$ Evaluate the left-hand side to get the answer.