How do you determine if the series $\sum\limits_{k=1}^\infty \left(1-\frac1k\right)^{k^2}$ converges? $$\sum_{k=1}^\infty \mathrm{(1-\frac{1}{k})}^{\mathrm{k}^{2}}$$
I tried using the limit comparison test with $$\sum_{k=1}^\infty \mathrm{(1-\frac{1}{k})}^{\mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of 
$\mathrm{e}^x$, but I don't know where else to start. Any suggestions?
 A: The root test works. Consider
$$\lim \sup \sqrt[k]{\left(1 - \frac{1}{k}\right)^{k^2}} = \lim \sup \left(1 - \frac{1}{k}\right)^k = e^{-1} < 1,$$
hence the series converges.
A: $\begin{array}\\
(1-\frac1{k})^{k^2}
&=(\frac{k-1}{k})^{k^2}\\
&=\dfrac1{(\frac{k}{k-1})^{k^2}}\\
&=\dfrac1{(1+\frac{1}{k-1})^{k^2}}\\
&=\dfrac1{((1+\frac{1}{k-1})^{k})^k}\\
&<\dfrac1{(1+\frac{k}{k-1})^k}
\qquad\text{by Bernoulli}\\
&=\dfrac1{(\frac{2k-1}{k-1})^k}\\
&<\dfrac1{(\frac{2k-2}{k-1})^k}\\
&=\dfrac1{2^k}\\
\end{array}
$
and the sum of this converges.
A: HINT:
Note that $$\left( 1-\frac1k \right)^k\le e^{-1}$$
Can you finish?
A: The ratio test is also interesting
$$a_k=\left(1-\frac{1}{k}\right)^{k^2}\implies \log(a_k)=k ^2 \log\left(1-\frac{1}{k}\right)$$
$$\log(a_{k+1})-\log(a_k)=(k+1) ^2 \log\left(1-\frac{1}{k+1}\right)-k ^2 \log\left(1-\frac{1}{k}\right)$$
Develop as a Taylor series for large values of $k$ to get
$$\log(a_{k+1})-\log(a_k)=-1+\frac{1}{3 k^2}+O(\left(\frac{1}{k^3}\right)$$
Continue with Taylor
$$\frac{a_{k+1}}{a_k}=e^{\log(a_{k+1})-\log(a_k)}=\frac 1 e\left(1+\frac{1}{3 k^2}+O\left(\frac{1}{k^3}\right)\right)\to \frac 1 e$$
