Limit of a sequence

Let an:=∑i=0nnii! a_n:=\sum_{i=0}^n \frac{n^i}{i!}. Find lim⁡n→+∞anen\lim_{n \to +\infty} \frac{a_n}{e^n}

#Limits #MathProblem #Math

Note by Lorenzo Sarnataro
7 years, 9 months ago

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Comments

lim⁡n→∞anen!\displaystyle \lim_{n \to \infty} \frac{a_{n}}{e^{n!}}

= lim⁡n→∞1+n1!+n22!+⋯+nn−1(n−1)!+nnn!1+n1!+n22!+⋯+nn−1(n−1)!+nnn!+nn+1(n+1)!+…\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, ⋯<nn−2(n−2)!<nn−1(n−1)!=nnn!>nn+1(n+1)!>nn+2(n+2)!>…\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 nnn!\frac{n^n}{n!} and then starts decreasing.

Suggest something to be done further.

kushagraa aggarwal - 7 years, 9 months ago

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Is the denominator en!e^{n!} or ene^{n}??

Krishna Jha - 7 years, 9 months ago

I am not of much help here but evaluating this with Wolfram | Alpha gives a very strange answer. Click here.

Pranav Arora - 7 years, 9 months ago

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The denominator is n! n! , since Γ(n)=(n−1)! \Gamma(n) = (n-1)! (for integer n 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 1/2 .

George Williams - 7 years, 9 months ago

a(n) is also series expansion of e^n ..... hence following limit becomes 1.

Anurag Sharma - 7 years, 9 months ago
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