sequences and series - Does $sum_{n=1}^infty (frac{1}{n}-1)^{n^2}$ converge?

I am trying to decide if $$\sum_{n=1}^\infty \left(\frac{1}{n}-1\right)^{n^2}$$ converges. By the alternating series test, as far as I can see, the series converges. This is also true by the root test. In both cases I assume that $$\left|\left(\frac{1}{n}-1\right)^{n^2}\right| = \left(1-\frac{1}{n}\right)^{n^2}$$ and I can't see anything wrong with that. What makes me unsure is that Wolfram Alpha says that the sum does not converge.

Wolfram Alpha uses the limit test, which I am not able to complete:
$$\lim_{n\to\infty} a_n = \lim_{n\to\infty}\left(\frac{1}{n}-1\right)^{n^2} = \lim_{n\to\infty} e^{n^2 \ln{\frac{1-n}{n}}}$$
The problem is the logarithm which is not defined for any $n$, and I am not aware of another way to compute that limit.

So does this series converge?

Edit: When I ask Wolfram Alpha to evaluate the sum, it says that the seires do not converge. But it also says that "computation timed out". Maybe this could be the reason for why it's wrong.

Answer

$$\left| \left( \frac{1}{n} - 1\right)^{n^2} \right| = \left| \left( \frac{1}{n} - 1\right) \right|^{n^2} = \left( 1 - \frac{1}{n}\right)^{n^2}$$
as you correctly noted. And we have
$$\limsup_{n\to\infty} \sqrt[n]{\left( 1 - \frac{1}{n}\right)^{n^2}} = \limsup_{n\to\infty} \left( 1 - \frac{1}{n}\right)^n = \frac{1}{e} < 1$$
so this series converges.