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Question Number 129997 by mohammad17 last updated on 21/Jan/21

prove that [(cos(n)+sin(n))/n] converge sequence

$${prove}\:{that}\:\left[\left({cos}\left({n}\right)+{sin}\left({n}\right)\right)/{n}\right]\:{converge}\:{sequence} \\ $$

Answered by Dwaipayan Shikari last updated on 21/Jan/21

Σ_(n=1) ^∞ ((cosn+sinn)/n)=(1/2)Σ_(n=1) ^∞ (e^(in) /n)+(e^(−in) /n)+(1/(2i))Σ_(n=1) ^∞ (e^(in) /n)−(e^(−in) /n) =−(1/2)(log(1−2cos1+1)+log(((1−e^i )/(1−e^(−i) ))) =(1/2)log((1/(4sin^2 (1/2))))+(π/2)−(1/2)=(1/2)(π−1−log(4sin^2 (1/2)))∼1.11

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cosn}+{sinn}}{{n}}=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{in}} }{{n}}+\frac{{e}^{−{in}} }{{n}}+\frac{\mathrm{1}}{\mathrm{2}{i}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{in}} }{{n}}−\frac{{e}^{−{in}} }{{n}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left({log}\left(\mathrm{1}−\mathrm{2}{cos}\mathrm{1}+\mathrm{1}\right)+{log}\left(\frac{\mathrm{1}−{e}^{{i}} }{\mathrm{1}−{e}^{−{i}} }\right)\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{\mathrm{1}}{\mathrm{4}{sin}^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}}\right)+\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi−\mathrm{1}−{log}\left(\mathrm{4}{sin}^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}\right)\right)\sim\mathrm{1}.\mathrm{11} \\ $$

Commented by mohammad17 last updated on 21/Jan/21

thank you sir but i want this by analyetic function

$${thank}\:{you}\:{sir}\:{but}\:{i}\:{want}\:{this}\:{by}\:{analyetic}\:{function} \\ $$

Answered by mathmax by abdo last updated on 21/Jan/21

let u_n =((cos(n)+sin(n))/n) ⇒u_n =(((√2)cos(n−(π/4)))/n) ⇒Σ_(n=1) ^∞ u_n =(√2)Σ_(n=1) ^∞ ((cos(n−(π/4)))/n)=(√2)Σ_(n=1) ^∞ α_n .β_n α_n decrease to and we can determine m>0 / ∣Σ_(k=1) ^n β_k ∣≤m so the convergence is assured by abel dirichlet theorem..

$$\mathrm{let}\:\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{cos}\left(\mathrm{n}\right)+\mathrm{sin}\left(\mathrm{n}\right)}{\mathrm{n}}\:\Rightarrow\mathrm{u}_{\mathrm{n}} =\frac{\sqrt{\mathrm{2}}\mathrm{cos}\left(\mathrm{n}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{n}} \\ $$$$\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{u}_{\mathrm{n}} =\sqrt{\mathrm{2}}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{n}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{n}}=\sqrt{\mathrm{2}}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\alpha_{\mathrm{n}} .\beta_{\mathrm{n}} \\ $$$$\alpha_{\mathrm{n}} \:\mathrm{decrease}\:\mathrm{to}\:\:\mathrm{and}\:\mathrm{we}\:\mathrm{can}\:\mathrm{determine}\:\mathrm{m}>\mathrm{0}\:/ \\ $$$$\mid\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\beta_{\mathrm{k}} \mid\leqslant\mathrm{m}\:\:\:\mathrm{so}\:\mathrm{the}\:\mathrm{convergence}\:\mathrm{is}\:\mathrm{assured}\:\mathrm{by}\:\mathrm{abel}\:\mathrm{dirichlet}\: \\ $$$$\mathrm{theorem}.. \\ $$

Commented by mathmax by abdo last updated on 21/Jan/21

α_n decrease to 0

$$\alpha_{\mathrm{n}} \:\mathrm{decrease}\:\mathrm{to}\:\mathrm{0} \\ $$

Commented by mohammad17 last updated on 21/Jan/21

but sir this is sequence not series and my teacher sayd me it want the solution by ∀∈>0 ∃k∈N Such that ∣x_n −x_0 ∣<∈ ∀n>k can help me?sir

$${but}\:{sir}\:{this}\:{is}\:{sequence}\:{not}\:{series}\:{and}\:{my}\: \\ $$$${teacher}\:{sayd}\:{me}\:{it}\:{want}\:{the}\:{solution}\:{by} \\ $$$$ \\ $$$$\forall\in>\mathrm{0}\:\exists{k}\in{N}\:{Such}\:{that}\:\mid{x}_{{n}} −{x}_{\mathrm{0}} \mid<\in\:\forall{n}>{k} \\ $$$$ \\ $$$${can}\:{help}\:{me}?{sir} \\ $$

Commented by mathmax by abdo last updated on 22/Jan/21

your teacher want to turn you to stone era ...no need to use ξ look we have ((cosn +sinn)/n)=(((√2)cos(n−(π/4)))/n) ⇒ ∣((cos(n)+sin(n))/n)∣≤((√2)/n)→0 ⇒lim_(n→+∞) ((cosn +sinn)/n)=0

$$\mathrm{your}\:\mathrm{teacher}\:\mathrm{want}\:\mathrm{to}\:\mathrm{turn}\:\mathrm{you}\:\mathrm{to}\:\mathrm{stone}\:\mathrm{era}\:...\mathrm{no}\:\mathrm{need}\:\mathrm{to}\:\mathrm{use}\:\xi \\ $$$$\mathrm{look}\:\mathrm{we}\:\mathrm{have}\:\:\frac{\mathrm{cosn}\:+\mathrm{sinn}}{\mathrm{n}}=\frac{\sqrt{\mathrm{2}}\mathrm{cos}\left(\mathrm{n}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{n}}\:\Rightarrow \\ $$$$\mid\frac{\mathrm{cos}\left(\mathrm{n}\right)+\mathrm{sin}\left(\mathrm{n}\right)}{\mathrm{n}}\mid\leqslant\frac{\sqrt{\mathrm{2}}}{\mathrm{n}}\rightarrow\mathrm{0}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{cosn}\:+\mathrm{sinn}}{\mathrm{n}}=\mathrm{0} \\ $$

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