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

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  • $\begingroup$ This is susceptible to the same approach as my answer to a different question. $\endgroup$
    – user296602
    Feb 23, 2019 at 3:16
  • $\begingroup$ Do you know what $\left(1-\frac{1}{k}\right)^{k}$ converges to? $\endgroup$
    – JavaMan
    Feb 23, 2019 at 3:22
  • $\begingroup$ I cannot believe how horrible my intuition is with this stuff especially given how old I am. $\endgroup$
    – Randall
    Feb 23, 2019 at 3:52
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    $\begingroup$ See also: Convergence of $\sum _{k=1}^\infty (1-\frac{1}{k})^{k^2}$ $\endgroup$ Feb 23, 2019 at 6:18

4 Answers 4

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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.

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$\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.

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HINT:

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

Can you finish?

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

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