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Question Number 46187 by annika0209 last updated on 22/Oct/18
Σ_(n=1) ^∞ (1/(n×n^(1/n) ))=? please help me!!!!
$$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}×\mathrm{n}^{\frac{\mathrm{1}}{\mathrm{n}}} }=?\:\:\:\:\:\:\:\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}!!!! \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 22/Oct/18
let S_n =Σ_(k=1) ^n (1/(k k^(1/k) )) the sequence u_k =(1/(k.k^(1/k) )) is decreasing ⇒ ∫_k ^(k+1) (dt/(t t^(1/t) )) ≤ f(k) ≤ ∫_(k−1) ^k (dt/(t t^(1/t) )) ⇒ Σ_(k=2) ^n ∫_k ^(k+1) (dt/(t .t^(1/t) ))≤Σ_(k=2) ^n (1/(k.k^(1/k) )) ≤ Σ_(k=2) ^n ∫_(k−1) ^k (dt/(tt^(1/t) )) ⇒ ∫_2 ^(n+1) (dt/(t.t^(1/t) )) ≤ S_n −1 ≤ ∫_1 ^n (dt/(t t^(1/t) )) ⇒1+ ∫_2 ^(+∞) (dt/(t.t^(1/t) )) ≤lim_(n→+∞) S_n ≤1+∫_1 ^(+∞) (dt/(t.t^(1/t) )) but ∫_1 ^(+∞) (dt/(t t^(1/t) )) =∫_1 ^(+∞) (dt/t^(1+(1/t)) ) =∫_1 ^(+∞) (dt/e^((1+(1/t))ln(t)) ) = ∫_1 ^(+∞) e^(−(1+(1/t))ln(t)) dt changement t =(1/x) give ∫_1 ^(+∞) e^(−(1+(1/t))ln(t)) dt = −∫_0 ^1 e^((1+x)ln(x)) (−(dx/x^2 ))= ∫_0 ^1 (e^((1+x)ln(x)) /x^2 )dx = ∫_0 ^1 ((x x^x )/x^2 )dx = ∫_0 ^1 x^(x−1) dx ....be continued...
$${let}\:\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}\:{k}^{\frac{\mathrm{1}}{{k}}} }\:\:\:{the}\:{sequence}\:{u}_{{k}} =\frac{\mathrm{1}}{{k}.{k}^{\frac{\mathrm{1}}{{k}}} }\:{is}\:{decreasing}\:\Rightarrow \\ $$$$\:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\frac{{dt}}{{t}\:{t}^{\frac{\mathrm{1}}{{t}}} }\:\leqslant\:{f}\left({k}\right)\:\leqslant\:\int_{{k}−\mathrm{1}} ^{{k}} \:\frac{{dt}}{{t}\:{t}^{\frac{\mathrm{1}}{{t}}} }\:\Rightarrow\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\frac{{dt}}{{t}\:.{t}^{\frac{\mathrm{1}}{{t}}} }\leqslant\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}.{k}^{\frac{\mathrm{1}}{{k}}} }\:\leqslant\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\int_{{k}−\mathrm{1}} ^{{k}} \:\frac{{dt}}{{tt}^{\frac{\mathrm{1}}{{t}}} }\:\Rightarrow \\ $$$$\int_{\mathrm{2}} ^{{n}+\mathrm{1}} \:\:\:\frac{{dt}}{{t}.{t}^{\frac{\mathrm{1}}{{t}}} }\:\leqslant\:{S}_{{n}} −\mathrm{1}\:\leqslant\:\int_{\mathrm{1}} ^{{n}} \:\:\:\frac{{dt}}{{t}\:{t}^{\frac{\mathrm{1}}{{t}}} }\:\Rightarrow\mathrm{1}+\:\int_{\mathrm{2}} ^{+\infty} \:\:\frac{{dt}}{{t}.{t}^{\frac{\mathrm{1}}{{t}}} }\:\leqslant{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \:\leqslant\mathrm{1}+\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dt}}{{t}.{t}^{\frac{\mathrm{1}}{{t}}} } \\ $$$${but}\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{dt}}{{t}\:{t}^{\frac{\mathrm{1}}{{t}}} }\:=\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dt}}{{t}^{\mathrm{1}+\frac{\mathrm{1}}{{t}}} }\:=\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{dt}}{{e}^{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right){ln}\left({t}\right)} } \\ $$$$=\:\int_{\mathrm{1}} ^{+\infty} \:\:{e}^{−\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right){ln}\left({t}\right)} {dt}\:{changement}\:{t}\:=\frac{\mathrm{1}}{{x}}\:{give} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\:{e}^{−\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right){ln}\left({t}\right)} {dt}\:=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{e}^{\left(\mathrm{1}+{x}\right){ln}\left({x}\right)} \:\left(−\frac{{dx}}{{x}^{\mathrm{2}} }\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{e}^{\left(\mathrm{1}+{x}\right){ln}\left({x}\right)} }{{x}^{\mathrm{2}} }{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{x}\:{x}^{{x}} }{{x}^{\mathrm{2}} }{dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{{x}−\mathrm{1}} \:{dx}\:\:….{be}\:{continued}… \\ $$

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