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Question Number 145960 by puissant last updated on 10/Jul/21

Answered by mathmax by abdo last updated on 10/Jul/21

R_n =Σ_(p=n+1) ^(2n) sin((1/p)) ⇒R_n =_(p−n=k) Σ_(k=1) ^n sin((1/(n+k))) we have sinx=x−(x^3 /6)+... ⇒x−(x^3 /6)≤sinx≤x ⇒ (1/(n+k))−(1/(6(n+k)^3 ))≤sin((1/(n+k)))≤(1/(n+k)) ⇒ Σ_(k=1) ^n (1/(n+k))−(1/6)Σ_(k=1) ^n (1/((n+k)^3 ))≤Σ_(k=1) ^n sin((1/(n+k)))≤Σ_(k=1) ^n (1/(n+k)) Σ_(k=1) ^n (1/(n+k))=(1/n)Σ_(k=1) ^n (1/(1+(k/n)))→∫_0 ^1 (dx/(1+x))=log2 Σ_(k=1) ^n (1/((n+k)^3 )) we k≥1 ⇒n+k≥n+1 ⇒(1/((n+k)^3 ))≤(1/((n+1)^3 )) ⇒Σ_(k=1) ^n (1/((n+k)^3 ))≤(n/((n+1)^3 ))→0 (n→∞) ⇒lim_(n→+∞) R_n =log2

$$\mathrm{R}_{\mathrm{n}} =\sum_{\mathrm{p}=\mathrm{n}+\mathrm{1}} ^{\mathrm{2n}} \mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{p}}\right)\:\Rightarrow\mathrm{R}_{\mathrm{n}} =_{\mathrm{p}−\mathrm{n}=\mathrm{k}} \:\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\:\mathrm{sinx}=\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}+...\:\Rightarrow\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\leqslant\mathrm{sinx}\leqslant\mathrm{x}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{6}\left(\mathrm{n}+\mathrm{k}\right)^{\mathrm{3}} }\leqslant\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}\right)\leqslant\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}\:\Rightarrow \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{6}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{k}\right)^{\mathrm{3}} }\leqslant\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}\right)\leqslant\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}} \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}}=\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}}\rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}}=\mathrm{log2} \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{k}\right)^{\mathrm{3}} }\:\:\:\mathrm{we}\:\mathrm{k}\geqslant\mathrm{1}\:\Rightarrow\mathrm{n}+\mathrm{k}\geqslant\mathrm{n}+\mathrm{1}\:\Rightarrow\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{k}\right)^{\mathrm{3}} }\leqslant\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{k}\right)^{\mathrm{3}} }\leqslant\frac{\mathrm{n}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }\rightarrow\mathrm{0}\:\left(\mathrm{n}\rightarrow\infty\right)\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{R}_{\mathrm{n}} =\mathrm{log2} \\ $$

Commented by mathmax by abdo last updated on 10/Jul/21

you are welcome sir.

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir}. \\ $$

Commented by puissant last updated on 10/Jul/21

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by puissant last updated on 10/Jul/21

okey thanks

$$\mathrm{okey}\:\:\mathrm{thanks} \\ $$

Answered by mathmax by abdo last updated on 10/Jul/21

A_n =Σ_(p=n+1) ^(2n) (1/p^α ) ⇒A_n =_(p−n=k) Σ_(k=1) ^n (1/((n+k)^α )) n+k>n ⇒(1/((n+k)^α ))<(1/n^α ) ⇒A_n <(n/n^α )=(1/n^(α−1) ) but α−1>0 ⇒lim_(n→∞) (1/n^(α−1) )=0 ⇒lim_(n→+∞) A_n =0

$$\mathrm{A}_{\mathrm{n}} =\sum_{\mathrm{p}=\mathrm{n}+\mathrm{1}} ^{\mathrm{2n}} \:\frac{\mathrm{1}}{\mathrm{p}^{\alpha} }\:\Rightarrow\mathrm{A}_{\mathrm{n}} =_{\mathrm{p}−\mathrm{n}=\mathrm{k}} \:\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{k}\right)^{\alpha} } \\ $$$$\mathrm{n}+\mathrm{k}>\mathrm{n}\:\Rightarrow\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{k}\right)^{\alpha} }<\frac{\mathrm{1}}{\mathrm{n}^{\alpha} }\:\Rightarrow\mathrm{A}_{\mathrm{n}} <\frac{\mathrm{n}}{\mathrm{n}^{\alpha} }=\frac{\mathrm{1}}{\mathrm{n}^{\alpha−\mathrm{1}} } \\ $$$$\mathrm{but}\:\alpha−\mathrm{1}>\mathrm{0}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\alpha−\mathrm{1}} }=\mathrm{0}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{A}_{\mathrm{n}} =\mathrm{0} \\ $$

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