Is the series $\sum e^{an^2}(1-\frac{a}{n})^{n^3}$ convergent? I stumbled upon this sum: 
$$\sum e^{an^2} (1-\frac{a}{n})^{n^3}$$
 Wolfram Alpha tells me it is convergent but I can't find a convenient proof to use.
The $n$-th root test seems to be useful, since we get $e^{an}(1-\frac{a}{n})^{n^2}$ and apparently this tends to $e^{-\frac{a^2}{2}}<1$ as long as $a\neq 0$. I say apparently because Wolfram says so but I can't seem to prove it; passing to logarithms, I can't use the equivalent $\log(1+u)\sim u$ when $u\to 0$ since things cancel out.
In any case, my main question is: can you provide a proof for convergence that doesn't rely on the root test? My bonus question is, please help me compute the limit above.
 A: Main Answer
Using the power series for $\log(1-x)$, we have
$$
\begin{align}
n^3\log\left(1-\frac an\right)
&=n^3\left[-\frac an-\frac12\frac{a^2}{n^2}-\frac13\frac{a^3}{n^3}+O\left(\frac1{n^4}\right)\right]\\
&=-an^2-\frac12a^2n-\frac13a^3+O\left(\frac1n\right)
\end{align}
$$
Therefore,
$$
\begin{align}
e^{\large an^2}\left(1-\frac an\right)^{\large n^3}
&=\exp\left[-\frac12a^2n-\frac13a^3+O\left(\frac1n\right)\right]\\
&=O\left(e^{\large-\frac12a^2n}\right)
\end{align}
$$
which is summable for $a\ne0$.

Bonus Answer
Since
$$
\begin{align}
n^2\log\left(1-\frac an\right)
&=n^2\left[-\frac an-\frac12\frac{a^2}{n^2}+O\left(\frac1{n^3}\right)\right]\\
&=-an-\frac12a^2+O\left(\frac1n\right)
\end{align}
$$
Therefore,
$$
\lim_{n\to\infty}e^{an}\left(1-\frac an\right)^{\large n^2}=e^{\large-\frac12a^2}
$$
A: Our first test to try is the limit test, since that exponential looks suspiciously divergent to me.
$$
\begin{align}
&\lim_{n\to\infty}e^{an^2}\left(1-\frac a n\right)^{n^3}\\
=&\lim_{n\to\infty}\exp\left\{\log\left(e^{an^2}\left(1-\frac a n\right)^{n^3}\right)\right\}\\
=&\lim_{n\to\infty}\exp\left\{an^2+n^3\log\left(1-\frac a n\right)\right\}
\end{align}
$$
As $n$ becomes large, $\frac a n$ becomes small.  We can take a series approximation to the logarithm and (after expansion) keep the dominant term:
$$
\begin{align}
=&\lim_{n\to\infty}\exp\left\{an^2+n^3\left(-\frac a n-\frac{a^2}{2n^2}-\frac{a^3}{3n^3}-\frac{a^4}{4n^4}-\cdots\right)\right\}\\
=&\lim_{n\to\infty}\exp\left\{an^2-an^2-\frac{a^2n}{2}-\frac{a^3}{3}-\frac{a^4}{4n}-\cdots\right\}\\
=&\lim_{n\to\infty}\exp\left\{-\frac{a^2n}{2}-\cdots\right\}\\
=e^\frac{a^2}2&\lim_{n\to\infty}e^{-n}=0
\end{align}
$$
Note that the last line holds only if $a\neq0$, otherwise every term in the series is $1$.  Therefore the sum diverges at $a=0$.  I bet that the convergence has something to do with the fact that $\lim_{n\to\infty}\left(1+\frac 1n\right)^n=e$.
Now we can move on to the ratio test.
$$
\begin{align}
&\lim_{n\to\infty}\frac{e^{a(n+1)^2}\left(1-\frac a{n+1}\right)^{(n+1)^3}}{e^{an^2}\left(1-\frac a n\right)^{n^3}}\\
=&\lim_{n\to\infty}\exp\left\{a(n+1)^2-an^2+(n+1)^3\log\left(1-\frac a{n+1}\right)-n^3\log\left(1-\frac a n\right)\right\}\\
=&\lim_{n\to\infty}\exp\left\{a+2an-an^2-\frac{4+a}2an-\cdots+an^2+\frac a2an+\cdots\right\}\\
=&\lim_{n\to\infty}\exp\left\{a+2an-\frac 42an-\frac 12(2+a)a\cdots\right\}\\
=&\lim_{n\to\infty}\exp\left\{-\frac{a^2}2+\cdots\right\}=e^{-a^2/2}
\end{align}
$$
(I'll admit I had Mathematica calculate those series!)  Since we know from before that $a\neq0$, $-\frac{a^2}2<0$ (since $a^2$ is positive for all real nonzero $a$) and therefore its exponential must be less than $1$.
Therefore by the ratio test the sum converges for all $a\neq0$ and by the limit test the sum diverges for $a=0$.
Now, for extra credit, I'll do the root test.  You correctly calculated the form of the limit:
$$
\begin{align}
&\lim_{n\to\infty}\sqrt[n]{e^{an^2}\left(1-\frac a n\right)^{n^3}}\\
=&\lim_{n\to\infty}\exp\left\{\frac 1n\left(an^2+n^3\log\left(1-\frac a n\right)\right)\right\}\\
=&\lim_{n\to\infty}\exp\left\{an+n^2\log\left(1-\frac a n\right)\right\}
\end{align}
$$
However, you got stuck when the first term of the $\log$ series expansion canceled.  The full series expansion (which I used above) is
$$
\log(1+u)=u+\frac{u^2}2+\frac{u^3}3+\cdots+\frac{u^n}n+\cdots
$$
I'll use the first three terms to evaluate our limit:
$$
\begin{align}
=&\lim_{n\to\infty}\exp\left\{an+n^2\left(-\frac a n-\frac{a^2}{2n^2}-\frac{a^3}{3n^3}-\cdots\right)\right\}\\
=&\lim_{n\to\infty}\exp\left\{an-an-\frac{a^2}2-\frac{a^3}{3n}-\cdots\right\}\\
=&\lim_{n\to\infty}\exp\left\{-\frac{a^2}2-\frac{a^3}{3n}-\cdots\right\}
\end{align}
$$
The first term canceled, but the rest of the terms stay.  Note that of the remaining terms, the second has $n$ in the denominator, so in the limit it tends to $0$.  In fact, since each one of the terms following has one more power of $n$ in the denominator they all tend to $0$.  This leaves us with our final expression:
$$
=\lim_{n\to\infty}\exp\left\{-\frac{a^2}2-\cdots\right\}=e^{-a^2/2}
$$
Which, as before, is less than $1$ when $a\neq 0$, meaning that by the root test the series converges when $a\neq 0$.
