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Question Number 169166 by infinityaction last updated on 25/Apr/22
find for arbitrary α≥0 the limit lim_(n→+∞) (((1^α +2^α +3^α +......n^α )/n^α )−(n/(1+α)))
$$\:\:\:\:\:\:\:\:\mathrm{find}\:\mathrm{for}\:\mathrm{arbitrary}\:\alpha\geqslant\mathrm{0}\:\mathrm{the}\:\mathrm{limit}\: \\ $$$$\:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow+\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}^{\alpha} +\mathrm{2}^{\alpha} +\mathrm{3}^{\alpha} +……\mathrm{n}^{\alpha} }{\mathrm{n}^{\alpha} }−\frac{\mathrm{n}}{\mathrm{1}+\alpha}\right) \\ $$
Commented by safojontoshtemirov last updated on 25/Apr/22
(1/2)
$$\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by infinityaction last updated on 25/Apr/22
sir solution
$${sir}\:{solution} \\ $$
Commented by infinityaction last updated on 25/Apr/22
sir dont put any value of α
$${sir}\:{dont}\:{put}\:{any}\:{value}\:{of}\:\alpha \\ $$
Commented by infinityaction last updated on 25/Apr/22
i want solution in general form
$${i}\:{want}\:{solution}\:{in}\:{general}\:{form} \\ $$
Answered by aleks041103 last updated on 25/Apr/22
∫_1 ^(n+1) x^α dx=(((n+1)^(α+1) −1)/(α+1))≈(Σ_(k=1) ^n k^α )+Σ_(k=1) ^n ((1/2)((k+1)^α −k^α ))= =(Σ_(k=1) ^n k^α )+(1/2)((n+1)^α −1) ⇒Σ_(k=1) ^n k^α ≈(((n+1)/(α+1))−(1/2))(n+1)^α +(1/2)−(1/(α+1)) ⇒lim_(n→∞) ((((n+1)/(α+1))−(1/2))(1+(1/n))^α −(n/(1+α))+((1/2)−(1/(α+1)))(1/n^α ))= =lim_(n→∞) ((1+(1/n))^α ((1/(α+1))−(1/2))+(n/(α+1))((1+(1/n))^α −1))= =(1/(α+1))−(1/2)+(α/(α+1))=(1/2) for α>0 for α=0 ⇒lim=((1+1+...+1)/1)−n=0 ⇒lim_(n→∞) (((1^α +...+n^α )/n^α )−(n/(α+1)))= { ((0, α=0)),((1/2, α>0)) :}
$$\int_{\mathrm{1}} ^{{n}+\mathrm{1}} {x}^{\alpha} {dx}=\frac{\left({n}+\mathrm{1}\right)^{\alpha+\mathrm{1}} −\mathrm{1}}{\alpha+\mathrm{1}}\approx\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\alpha} \right)+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\left({k}+\mathrm{1}\right)^{\alpha} −{k}^{\alpha} \right)\right)= \\ $$$$=\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\alpha} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\left({n}+\mathrm{1}\right)^{\alpha} −\mathrm{1}\right) \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\alpha} \approx\left(\frac{{n}+\mathrm{1}}{\alpha+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({n}+\mathrm{1}\right)^{\alpha} +\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\alpha+\mathrm{1}} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}}\left(\left(\frac{{n}+\mathrm{1}}{\alpha+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{\alpha} −\frac{{n}}{\mathrm{1}+\alpha}+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\alpha+\mathrm{1}}\right)\frac{\mathrm{1}}{{n}^{\alpha} }\right)= \\ $$$$=\underset{{n}\rightarrow\infty} {{lim}}\left(\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{\alpha} \left(\frac{\mathrm{1}}{\alpha+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{{n}}{\alpha+\mathrm{1}}\left(\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{\alpha} −\mathrm{1}\right)\right)= \\ $$$$=\frac{\mathrm{1}}{\alpha+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\alpha}{\alpha+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\:{for}\:\alpha>\mathrm{0} \\ $$$${for}\:\alpha=\mathrm{0} \\ $$$$\Rightarrow{lim}=\frac{\mathrm{1}+\mathrm{1}+…+\mathrm{1}}{\mathrm{1}}−{n}=\mathrm{0} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}}\left(\frac{\mathrm{1}^{\alpha} +…+{n}^{\alpha} }{{n}^{\alpha} }−\frac{{n}}{\alpha+\mathrm{1}}\right)=\begin{cases}{\mathrm{0},\:\alpha=\mathrm{0}}\\{\mathrm{1}/\mathrm{2},\:\alpha>\mathrm{0}}\end{cases} \\ $$
Commented by infinityaction last updated on 25/Apr/22
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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