Finding closed form expression for $ \sum_{n=1}^{\infty} n\left( e^{-n(n-1)} - e^{-n(n+1)}\right)$ I want to find a closed form expression for:
$$ \sum_{n=1}^{\infty} n\left( e^{-n(n-1)} - e^{-n(n+1)}\right)$$
This isn't a homework problem, more of something that emerged as a side problem from other work. This expression will converge and I was trying to find a closed form expression for the value but I'm kind of struggling. 
Any help?
 A: First notice that
$$
S=\sum_{n=1}^{\infty}n(e^{-n(n-1)} -e^{-n(n+1)}) = \sum_{n=1}^{\infty}ne^{-n(n-1)}  - \sum_{k=2}^{\infty}(k-1)e^{-k(k-1)} 
$$
where we made $n=k-1$, hence we get
$$
S = 1 + \sum_{n=2}^{\infty}e^{-n^2 +n}
$$
This last sum can be evaluated using the Jacobi theta function $\vartheta$. Finally we obtain the closed form
$$
S = \frac{1}{2} e^{1/4} \vartheta_2 (0, \frac{1}{e}) = 1.137820181\cdots
$$
A: $$ n\left( e^{-n(n-1)} - e^{-n(n+1)}\right) = n\left( \frac{1}{e^{n(n-1)}} - \frac{1}{e^{n(n+1)}} \right) = n\left( \frac{e^{n(n+1)} - e^{n(n-1)}}{e^{2n^2}} \right) = \frac{n\left(e^{n} - e^{-n} \right)}{e^{n^2}}$$
The series of that general term converges using the $n^\alpha$ test, which states that if the limit of $n^{\alpha}$ multiplied by the general term of a nonnegative series is zero, and $\alpha >1$, then the series converges.
We can multiply our series by any $n^{\alpha}$ we wish. $n^{100}$ for instance, and we'll still have a zero limit. 
A: $\left|n\left(e^{-n(n-1)}-e^{-n(n+1)}\right)\right|=\left|n\left(\dfrac{1}{e^{n(n-1)}}-\dfrac{1}{e^{n(n+1)}}\right)\right|=n\cdot\dfrac{e^{n(n+1)}-e^{n(n-1)}}{e^{2n^2}}<\dfrac{2ne^{n(n+1)}}{e^{2n^2}}=2ne^{n-n^2}=\dfrac{2n}{e^{n^2-n}}< \dfrac{2n}{1+(n^2-n)+\dfrac{(n^2-n)^2}{2}}< \dfrac{4n}{(n^2-n)^2}<\dfrac{16}{n^3}$. Thus the series converges absolutely by comparison test with the convergent $p$ series with $p = 3$. I might answer the "wrong" question, but seem "satisfied" by my own answer....
