# Find the sum of the series $\sum_{n = 1}^{\infty} n^2e^{-nx^2 - nx}$

I can't find a way to calculate the sum of the following serie: $$\sum_{n=1}^{\infty} n^2e^{-nx^2-nx}$$

I tried everything, but I cannot figure it out ... can someone help me by finding a way? Many thanks... :"(

## 1 Answer

Putting $\mathrm e^{-(x^2+x)}=z$ and $|z|<1$ $$\sum_{n=1}^{\infty} n^2\mathrm e^{-n(x^2+x)}=\sum_{n=1}^{\infty} n^2z^n\tag 1$$ We know that the power series for:

$$\sum_{n=1}^\infty nz^n = \frac{z}{(1-z)^2} \qquad |z| <1$$

In order to find the power series for (1) we differentiate

$$\left(\sum_{n=1}^\infty nz^n\right)' = \sum_{n=1}^\infty n^2z^{n-1} = -\frac{z+1}{{(z-1)}^3} \qquad |z| <1$$

Finally, we multiply by $z$:

$$z\sum_{n=1}^\infty n^2z^{n-1} = \sum_{n=1}^\infty n^2z^{n} = -\frac{z(z+1)}{{(z-1)}^3}$$

• Thank you so much, you saved my day ! Best best regards :') – Clausisus Pollas Oct 4 '17 at 5:45
• You should use $z=\mathrm e^{-x^2-x}$ – gammatester Oct 4 '17 at 8:44
• Minor quibble: If $-1\lt x\lt0$, then $|z|\gt1$ and the series does not converge. – Barry Cipra Oct 4 '17 at 10:50