Saturday, August 18, 2018

sequences and series - limlimitsntoinftyfracnsqrt[n]n!=e


I don't know how to prove that limnnnn!=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 k2n can be estimated nkk!<nk(2n)k2nnnn!=nnn!12k2n and thus we find $$\begin{align}\frac{n^n}{n!}

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