Showing that: limn→∞n!nn=1e

Question

Showing that:  limn→∞n!nn=1e

Terry Ray · Accepted Answer

Have not found a way to rewrite your expression to get the desired result. However, here is a suggested approach.
Maybe rewrite the left-hand side as

\sqrt[n]{\frac{n!}{n^{n}}}

Take the logarithm. We get

\frac{1}{n} (\log (\frac{1}{n}) + \log (\frac{2}{n}) + \log (\frac{3}{n}) + \dots + \log (\frac{n}{n}))

Now think of the above sum as a Riemann sum for the not quite proper integral

\int_{0}^{1} \log x d x

alenahelenash · Accepted Answer

If ≥0, then the following inequality holds: limn→∞ an+1an≤limn→∞ ann≤limn→∞ ⊃an+1an Now let an=n!/nn. Then it follows that an+1an=(n+1)!(n+1)n+1n!nn=1(1+1n)n and hence limn→∞ infan+1an=limn→∞ supan+1an=1e This proves that ann→e−1

user_27qwe · Accepted Answer

I would like to use the following lemma: If limn→∞an=a and an&amp;gt;0 for all n, then we have limn→∞a1a2…ann=a  (1) Let an=(1+1n)n, then an&amp;gt;0 for all n and limn→∞an=e. Applying (*) we have e=limn→∞a1a2…ann=limn→∞(21)1(32)2…(n+1n)nn=limn→∞(n+1)nn!n(2)=limn→∞nn!n where we use (1) in the last equality to show that limn→∞1n!n=0 It follows from (2) that limn→∞n!nn=1e

Showing that: \lim_{n\to\infty}\frac{\root(n)(n!)}{n}=\frac{1}{e}

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