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Limit of a sequence

Let \( a_n:=\sum_{i=0}^n \frac{n^i}{i!}\). Find \[\lim_{n \to +\infty} \frac{a_n}{e^n}\]

Note by Lorenzo Sarnataro
3 years, 11 months ago

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\(\displaystyle \lim_{n \to \infty} \frac{a_{n}}{e^{n!}}\)

= \(\displaystyle \lim_{n \to \infty} \frac{1 + \frac{n}{1!} + \frac{n^2}{2!} + \dots + \frac{n^{n - 1}}{(n - 1)!} + \frac{n^n}{n!}}{1 + \frac{n}{1!} + \frac{n^2}{2!} + \dots + \frac{n^{n - 1}}{(n - 1)!} + \frac{n^n}{n!} + \frac{n^{n + 1}}{(n + 1)!} + \dots }\).

Since, \(\dots < \frac{n^{n - 2}}{(n - 2)!} < \frac{n^{n - 1}}{(n -1)!} = \frac{n^n}{n!} > \frac{n^{n + 1}}{(n+1)!} > \frac{n^{n + 2}}{(n+ 2)!} > \dots\) , we note that denominator is a sequence which increases till \(\frac{n^n}{n!} \) and then starts decreasing.

Suggest something to be done further. Kushagraa Aggarwal · 3 years, 11 months ago

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@Kushagraa Aggarwal Is the denominator \(e^{n!}\) or \(e^{n}\)?? Krishna Jha · 3 years, 11 months ago

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a(n) is also series expansion of e^n ..... hence following limit becomes 1. Anurag Sharma · 3 years, 11 months ago

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I am not of much help here but evaluating this with Wolfram | Alpha gives a very strange answer. Click here. Pranav Arora · 3 years, 11 months ago

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@Pranav Arora The denominator is \( n! \), since \( \Gamma(n) = (n-1)! \) (for integer \( n \)). The numerator is the upper incomplete gamma function, but it doesn't help much with evaluation in any case.

Numerically, the answer appears to be \( 1/2 \). George Williams · 3 years, 11 months ago

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