Determine whether $\sum_{n=1}^\infty (-1)^n \Big(1-\frac{2}{n}\Big)^n$ converges or diverges $$\sum_{n=1}^\infty (-1)^n \Big(1-\frac{2}{n}\Big)^n$$
If these terms were all positive I would have no problem concluding that the series diverges since the limit would be $e^{-2}>0$. However, I don't understand how the divergence test works with $(-1)^n$ added in. Is this still divergent? Both the Ratio and Root tests are inconclusive, and Wolfram Alpha is scratching its head. When I put partial sums above 1000, it spits out imaginary numbers somehow. Yet it seems to stay around 1.

The series is divergent. Introduce partial sum:
$$S_k=\sum_{n=0}^k(-1)^n\left(1-\frac{2}{n}\right)^n$$
$$|S_k-S_{k-1}|=\left(1-\frac{2}{k}\right)^k$$
$$\lim_{k \to \infty}|S_k-S_{k-1}|=\lim_{k\to\infty}\left(1-\frac{2}{k}\right)^k=e^{-2}$$
So the difference between even and odd partial sums converges to some non-zero value. For big values of $k$, partial sum oscillates between two values that differ by $e^{-2}\approx0.135335.$
For example: $S_{2000}=1.05715$ but $S_{2001}=0.921953$ and the difference is almost exactly $e^{-2}$.
By applying the same argument you can conclude that for any convergent series (including those alternating), $a_n\to0$ is the necessary condition.
