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A computer algebra system told me that \begin{equation} \lim_{n \to \infty} \left( \frac{1}{n} \right)^{1/n!} = 1 \end{equation} How can I show this? I tried applying the exponential and logarithm to see that this is equal to \begin{equation} \exp \lim_{n \to \infty} \left( \frac{1}{n!} \log \frac{1}{n} \right) \end{equation} But I'm not sure how to proceed.

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    Do you know $\lim_{x\to0}x^x=1$ ? – 2017-02-05
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    Yes, I can do that using the above method and then applying L'Hopital's rule. How does that help me? – 2017-02-05
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    Then it equals $$\left(\lim_{n\to\infty}\left(\frac{1}{n}\right)^{\frac{1}{n}}\right)^{\left(\displaystyle\lim_{n\to\infty}\frac{1}{(n-1)!}\right)}=1^0=1$$ – 2017-02-05

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HINT:

For $n\ge 1 $, $n!\ge n$. Therefore,

$$0>-\frac{\log(n)}{n!}\ge -\frac{\log(n)}{n}$$

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    Go Green, Go White! – 2017-02-05
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You can solve this by checking radius of convergent. 1/n! Will become 0 faster than 1/n. So we can conclude that the whole limit will becone 1.

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    This does not constitute a rigorous proof. – 2017-02-06