Limit with root
- Tolaso J Kos
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Limit with root
Prove that: \( \displaystyle \lim \sqrt[n]{\frac{n!}{n^n}}=e^{-1} \).
Imagination is much more important than knowledge.
Re: Limit with root
Replied by ex-member aziiri:
Take the \(\ln\) to get : \[\frac{1}{n}\sum_{k=1}^{n} \ln \frac{k}{n} \to \int_0^1 \ln(x) \ \mathrm{d}x =-1\] There are other methods such as Stolz lemma : if \(\frac{a_{n+1}}{a_n} \to l\in \mathbb{R}\) then \(\sqrt[n]{a_n} \to l\) for positive sequence, applying it gives the result directly.
Take the \(\ln\) to get : \[\frac{1}{n}\sum_{k=1}^{n} \ln \frac{k}{n} \to \int_0^1 \ln(x) \ \mathrm{d}x =-1\] There are other methods such as Stolz lemma : if \(\frac{a_{n+1}}{a_n} \to l\in \mathbb{R}\) then \(\sqrt[n]{a_n} \to l\) for positive sequence, applying it gives the result directly.
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