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Question Number 1444 by 112358 last updated on 04/Aug/15

$$Evaluatethefollowingintegral.$$$${\int }_0^n\lfloor x{\rfloor }^{1/\lfloor x\rfloor !}dx(n\in \mathbb{N})$$$$Here\lfloor x\rfloor istheinteger-partofx$$$$e.g\lfloor 0.12\rfloor =0,\lfloor 5.896\rfloor =5$$$$$$

Answered by 123456 last updated on 05/Aug/15

$$f\left(x\right)=\lfloor x{\rfloor }^{1/\lfloor x\rfloor !}\equiv n^{1/n!},x=n+r,x\in {\mathbb{R}}_{+},n\in \mathbb{N},r\in [0,1)$$$${\mathrm{I}}_n=\underset{0}{\stackrel{n}{\int }}\lfloor x{\rfloor }^{1/\lfloor x\rfloor !}dx$$$${\mathrm{I}}_0=0$$$${\mathrm{I}}_n=\underset{0}{\stackrel{n}{\int }}\lfloor x{\rfloor }^{1/\lfloor x\rfloor !}dx=\underset{i=0}{\sum ^{n-1}}\underset{i}{\stackrel{i+1}{\int }}{i}^{1/i!}dx$$$$=\underset{i=0}{\sum ^{n-1}}{i}^{1/i!}$$

Commented by 123456 last updated on 06/Aug/15

$$\mathrm{L}=\underset{n\to +\mathrm{\infty }}{\lim }{n}^{1/n!}=?$$$$\mathrm{D}=\underset{n\to +\mathrm{\infty }}{\lim }{n}^{1/n}$$$$\ln \mathrm{D}=\underset{n\to +\mathrm{\infty }}{\lim }\frac{\ln n}{n}\stackrel{\mathrm{L}.\mathrm{H}}{=}\underset{n\to 0}{\lim }\frac{1}{n}=0$$$$\mathrm{D}=1$$$$\ln \mathrm{L}=\underset{n\to +\mathrm{\infty }}{\lim }\frac{\ln n}{n!}=?$$$$0<n^{1/n!}<n^{1/n}\mathrm{for}\mathrm{largd}\mathrm{enought}n$$$$0<\frac{\ln n}{n!}\stackrel{?}{<}\frac{\ln n}{n}$$$$1/n!>0$$$$1<n^{1/n!}<n^{1/n}\mathrm{for}\mathrm{large}\mathrm{enought}n$$$$\mathrm{L}=1$$

Commented by 123456 last updated on 05/Aug/15

$$\underset{n\to +\mathrm{\infty }}{\lim }{n}^{1/n!}\ne 0$$$$\underset{i=0}{\sum ^{+\mathrm{\infty }}}{i}^{1/i!}\mathrm{diverge}$$

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