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.
 A: If the series $\sum a_n$ is convergent then $a_n \to 0$. This holds for any series, not just positive terms series. So, in the case you present, since $\lim a_n \ne 0$, you can conclude that the series is divergent.
A: If it's convergent then $\left | a_{n} \right |\rightarrow 0$
$(1-\frac{2}{n})^{n}=((1+\frac{1}{-2^{-1}n})^{-2^{-1}n})^{2}\rightarrow e^{2}\neq 0$
and therefore isn't convergent.
A: 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.
A: Calculate
$$\lim_{n \to +\infty} (-1)^n \Big(1-\frac{2}{n}\Big)^n$$
with the two subsequences $n = 2k$ and $n=2k+1$.
Then you'll have
$$\lim_{k \to +\infty} (-1)^{2k} \Big(1-\frac{2}{2k}\Big)^{2k} =  \lim_{k \to +\infty} \Big(1-\frac{1}{k}\Big)^{2k} = e^{-2}   $$
$$\lim_{k \to +\infty} (-1)^{2k+1} \Big(1-\frac{2}{2k+1}\Big)^{2k+1} =
\lim_{k \to +\infty} -\Big(1-\frac{2}{2k+1}\Big)^{2k+1}=  
-e^{-2}   $$
Since the limit doesn't exist the series diverges.
