Convergence of $\sum _{k=1}^\infty (1-\frac{1}{k})^{k^2}$ Found the alternative form: $\sum _{k=1}^\infty ((1-\frac{1}{k})^{k})^k$.
Tried various criteria, no luck so far.
 A: It's possible to throw various convergence tests at the series, but here is how the authors might have thought about it when posing the problem.
If you don't know that $(1-\frac{1}{k})^k$ converges to $1/e$, the problem makes no sense.
If you do know that, then the problem is plainly derived from the relation to $e$, and it "should" resemble $\sum e^{-k}$, a geometric series that converges. But it is not necessarily easy to make this precise, because the sequence converging to $1/e$ is being raised to a variable power, $k$, and this could (hypothetically) re-separate the things that are getting closer together.
I remember, not necessarily correctly,  that the convergence of $(1-x/n)^n$ to $e^{-x}$ is from below, the sequence increasing with $n$, so that we would have $S \leq \sum e^{-k}$ termwise, and this would finish the problem.
If one wants not to rely on knowledge about how $(1-1/k)^k$ converges to $1/e$ and only use that it does converge to that limit, we know that for $k$ past some value $k_0$, $(1-\frac{1}{k})^k < \frac{1}{2}$, because $e > 2$.  And then the series will be a sum of a finite number of terms, plus an infinite tail bounded by a sum of $2^{-k}$ for $k \geq k_0$.
A: Use limit comparison test with $\displaystyle \sum_{k=1}^{\infty} \dfrac1{e^k}$.
A: Using Root Test: Given $\sum a_n$
$$\lim_{n \to \infty} {|a_n|}^{\frac{1}{n}} = L$$
If $L < 1$, $\sum a_n$ converges absolutely. If $L > 1$, $\sum a_n$ diverges. If $L = 1$, the test is inconclusive.  
So your series becomes 
$$\lim_{k \to \infty} |{(1 - \frac{1}{k})}^{k^2}|^{\frac{1}{k}} = \lim_{k \to \infty} {(1 - \frac{1}{k})}^k$$
Using 
$$\lim_{x \to \infty} {(1 + \frac{1}{x})}^x = e \implies \lim_{x \to \infty} {(1 - \frac{1}{x})}^x = e^{-1}$$
$$\lim_{k \to \infty} {(1 - \frac{1}{k})}^k = e^{-1}$$
Since the limit, $L (\frac{1}{e}) < 1$, the series converges absolutely $\implies$ it converges in the usual sense.
