How do you determine if the series $\sum\limits_{k=1}^\infty \left(1-\frac1k\right)^{k^2}$ converges?

Question

$$\sum_{k=1}^\infty \mathrm{(1-\frac{1}{k})}^{\mathrm{k}^{2}}$$ 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?

Answer 1

HINT:

Note that $$\left( 1-\frac1k \right)^k\le e^{-1}$$

Can you finish?

Answer 2

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.

Answer 3

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 $$\frac{a_{k+1}}{a_k}=e^{\log(a_{k+1})-\log(a_k)}=\frac 1 e\left(1+\frac{1}{3 k^2}+O\left(\frac{1}{k^3}\right)\right)\to \frac 1 e$$

Answer 4

$\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.