I don't know how to prove that limn→∞nn√n!=e. Are there other different (nontrivial) nice limit that gives e apart from this and the following ∞∑k=01k!=limn→∞(1+1n)n=e?
Answer
In the series for en=∞∑k=0nkk!, the nth and biggest(!) of the (throughout positve) summands is nnn!. On the other hand, all summands can be esimated as nkk!≤nnn! and especially those with k≥2n can be estimated nkk!<nk(2n)k−2n⋅nn⋅n!=nnn!⋅12k−2n and thus we find $$\begin{align}\frac{n^n}{n!}
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