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So I was working on this problem and I got diverge, since my answer was greater than 1. The Limit was > 1, using the root test.

$$\sum\limits_{n=4}^\infty (1 +\frac{1}{n})^{-n^2}$$

I ended up with

$(1+ \frac{1}{n})^{-n}$ as the limit $n$ approaches $\infty$

Using the root test. I would assume that would be diverging, but apparently it's converge. Let me know what I did wrong.

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Recall that if $\lim_{n\to\infty} a_n = L$, and $L \neq 0$, then $$\lim_{n\to\infty} \frac{1}{a_n} = \frac1L$$ We know that $$\lim_{n\to\infty} \left(1+\frac1n \right)^n = e$$ by L'Hôpital's rule. So $$\lim_{n\to\infty} \frac{1}{(1+1/n)^n} = 1/e < 1$$ So by the root test, the sum converges.

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  • $\begingroup$ Makes sense. Thank you! $\endgroup$ – Adan Vivero May 1 at 0:35
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The limit of $(1+\frac 1 n)^{-n}$ is $\frac 1 e$ which is less than $1$.

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Even the ratio test works. Consider $$a_n=\left(1+\frac{1}{n}\right)^{-n^2}\implies \log(a_n)=-n^2 \log\left(1+\frac{1}{n}\right)$$ $$\log(a_{n+1})-\log(a_n)=n^2 \log \left(1+\frac{1}{n}\right)-(n+1)^2 \log \left(1+\frac{1}{n+1}\right)$$ Now, for large $n$, use Taylor expansion to get $$\log(a_{n+1})-\log(a_n)=-1+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\frac{ a_{n+1}} {a_n }=e^{\log(a_{n+1})-\log(a_n)}=\frac 1 e\left(1+\frac{1}{3 n^2}\right)+O\left(\frac{1}{n^3}\right) =\frac 1 e+O\left(\frac{1}{n^2}\right)$$

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