Weird Limit + Integration Mix

Calculus Level 5

Evaluate

lim n n e n e 0 n e x d x \large \lim_{n\to\infty}\frac{\sqrt{n}}{e^n}e^{\large\int_{0}^{\infty}\lfloor\frac{n}{e^x}\rfloor dx}

Notation: \lfloor \cdot \rfloor denotes the floor function .


The answer is 0.398942280401.

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2 solutions

Chew-Seong Cheong
Oct 14, 2019

L = lim n n e n exp ( 0 n e x d x ) = lim n n e n exp ( 0 ln ( n n 1 ) ( n 1 ) d x + ln ( n n 1 ) ln ( n n 2 ) ( n 2 ) d x + + ln ( n 3 ) ln ( n 2 ) 2 d x + ln ( n 2 ) ln n 1 d x ) = lim n n e n exp ( k = 1 n 1 ln ( n k + 1 ) ln ( n k ) k d x ) = lim n n e n exp ( k = 2 n ( k 1 ) ln k k = 1 n 1 k ln k ) = lim n n e n exp ( k = 2 n k ln k k = 1 n 1 k ln k k = 2 n ln k ) = lim n n e n exp ( n ln n ln ( n ! ) ) = lim n n e n ( n n n ! ) By Stirling’s formula = lim n n e n ( n n 2 π n ( n e ) n ) = 1 2 π 0.399 \begin{aligned} L & = \lim_{n \to \infty} \frac {\sqrt n}{e^n} \exp \left(\int_0^\infty \left \lfloor \frac n{e^x} \right \rfloor dx \right) \\ & = \lim_{n \to \infty} \frac {\sqrt n}{e^n} \exp \left(\int_0^{\ln \left(\frac n{n-1}\right)} (n-1) \ dx + \int_{\ln \left(\frac n{n-1}\right)}^{\ln \left(\frac n{n-2}\right)} (n-2) \ dx + \cdots + \int_{\ln \left(\frac n3\right)}^{\ln \left(\frac n2 \right)} 2 \ dx + \int_{\ln \left(\frac n2 \right)}^{\ln n} 1 \ dx \right) \\ & = \lim_{n \to \infty} \frac {\sqrt n}{e^n} \exp \left(\sum_{k=1}^{n-1} \int_{\ln \left(\frac n{k+1} \right)}^{\ln \left(\frac nk \right)} k \ dx \right) \\ & = \lim_{n \to \infty} \frac {\sqrt n}{e^n} \exp \left(\blue{\sum_{k=2}^n (k-1) \ln k} - \sum_{k=1}^{n-1} k \ln k \right) \\ & = \lim_{n \to \infty} \frac {\sqrt n}{e^n} \exp \left(\blue{\sum_{k=2}^n k \ln k} - \sum_{k=1}^{n-1} k \ln k - \blue{\sum_{k=2}^n \ln k} \right) \\ & = \lim_{n \to \infty} \frac {\sqrt n}{e^n} \exp \left(n \ln n - \ln (n!) \right) \\ & = \lim_{n \to \infty} \frac {\sqrt n}{e^n} \left(\frac {n^n}{\blue{n!}} \right) \quad \quad \small \blue{\text{By Stirling's formula}} \\ & = \lim_{n \to \infty} \frac {\sqrt n}{e^n} \left(\frac {n^n}{\blue{\sqrt{2\pi n}\left(\frac ne \right)^n}} \right) \\ & = \frac 1{\sqrt{2\pi}} \approx \boxed{0.399} \end{aligned}


Reference: Stirling's formula

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