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

Evaluate the following integral. ∫_0 ^( n) ⌊x⌋^(1/⌊x⌋!) dx (n∈N) Here ⌊x⌋ is the integer−part of x e.g ⌊0.12⌋=0, ⌊5.896⌋=5

$${Evaluate}\:{the}\:{following}\:{integral}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:{n}} \lfloor{x}\rfloor^{\mathrm{1}/\lfloor{x}\rfloor!} {dx}\:\:\:\:\:\:\:\left({n}\in\mathbb{N}\right) \\ $$$${Here}\:\lfloor{x}\rfloor\:{is}\:{the}\:{integer}−{part}\:{of}\:{x} \\ $$$${e}.{g}\:\lfloor\mathrm{0}.\mathrm{12}\rfloor=\mathrm{0},\:\lfloor\mathrm{5}.\mathrm{896}\rfloor=\mathrm{5} \\ $$$$ \\ $$

Answered by 123456 last updated on 05/Aug/15

f(x)=⌊x⌋^(1/⌊x⌋!) ≡n^(1/n!) ,x=n+r,x∈R_+ ,n∈N,r∈[0,1) I_n =∫_0 ^n ⌊x⌋^(1/⌊x⌋!) dx I_0 =0 I_n =∫_0 ^n ⌊x⌋^(1/⌊x⌋!) dx=Σ_(i=0) ^(n−1) ∫_i ^(i+1) i^(1/i!) dx =Σ_(i=0) ^(n−1) i^(1/i!)

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

Commented by 123456 last updated on 06/Aug/15

L=lim_(n→+∞) n^(1/n!) =? D=lim_(n→+∞) n^(1/n) ln D=lim_(n→+∞) ((ln n)/n)=^(L.H) lim_(n→0) (1/n)=0 D=1 ln L=lim_(n→+∞) ((ln n)/(n!))=? 0<n^(1/n!) <n^(1/n) for largd enought n 0<((ln n)/(n!))<^? ((ln n)/n) 1/n!>0 1<n^(1/n!) <n^(1/n) for large enought n L=1

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

Commented by 123456 last updated on 05/Aug/15

lim_(n→+∞) n^(1/n!) ≠0 Σ_(i=0) ^(+∞) i^(1/i!) diverge

$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{n}^{\mathrm{1}/{n}!} \neq\mathrm{0} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{+\infty} {\sum}}{i}^{\mathrm{1}/{i}!} \:\mathrm{diverge} \\ $$

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