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Question Number 86113 by Tony Lin last updated on 27/Mar/20

(1)Determine the following if it is convergent or divergent Σ_(n=1) ^∞ ((sin(n))/n) (2)Σ_(n=1) ^∞ ((sin(n^p ))/n^p ), pεR,find the range of p when it is convergent

$$\left(\mathrm{1}\right){Determine}\:{the}\:{following} \\ $$$${if}\:{it}\:{is}\:{convergent}\:{or}\:{divergent} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({n}\right)}{{n}} \\ $$$$\left(\mathrm{2}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({n}^{{p}} \right)}{{n}^{{p}} },\:{p}\epsilon\mathbb{R},{find}\:{the}\:{range}\: \\ $$$${of}\:{p}\:{when}\:{it}\:{is}\:{convergent} \\ $$

Commented by Serlea last updated on 27/Mar/20

Σ_(n=1) ^∞ ((sin(n))/n) let U_n =((sin(n))/((n))) =((1/n))sin(n) V_n =(n/n)=1 lim_(n→∞) (U_n /V_n )=lim_(n→∞) ((sin(n))/n)=1 ΣU_n and ΣV_(n ) but converges and diverges It′s diverent from the P−series test (p=0<1) 2) Σ((sin(n^p ))/n^p ) U_(n ) =((sin(n^p ))/n^p ) V_n =(n^P /n^p )=1 Lim_(n→∞) (U_n /V_n )=lim_(n→∞ ) ((sin(n^p ))/n^p ) let t=n^p =1 P=1 Convergent when p≥1

$$ \\ $$$$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({n}\right)}{{n}} \\ $$$$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{sin}\left(\mathrm{n}\right)}{\left(\mathrm{n}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{1}}{\mathrm{n}}\right)\mathrm{sin}\left(\mathrm{n}\right) \\ $$$$\mathrm{V}_{\mathrm{n}} =\frac{\mathrm{n}}{\mathrm{n}}=\mathrm{1} \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{U}_{\mathrm{n}} }{\mathrm{V}_{\mathrm{n}} }=\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{sin}\left(\mathrm{n}\right)}{\mathrm{n}}=\mathrm{1} \\ $$$$\Sigma\mathrm{U}_{\mathrm{n}} \:\mathrm{and}\:\Sigma\mathrm{V}_{\mathrm{n}\:} \:\mathrm{but}\:\mathrm{converges}\:\mathrm{and}\:\mathrm{diverges} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{diverent}\:\mathrm{from}\:\mathrm{the}\:\mathrm{P}−\mathrm{series}\:\mathrm{test} \\ $$$$\left(\mathrm{p}=\mathrm{0}<\mathrm{1}\right) \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\Sigma\frac{\mathrm{sin}\left(\mathrm{n}^{\mathrm{p}} \right)}{\mathrm{n}^{\mathrm{p}} } \\ $$$$\mathrm{U}_{\mathrm{n}\:} =\frac{\mathrm{sin}\left(\mathrm{n}^{\mathrm{p}} \right)}{\mathrm{n}^{\mathrm{p}} } \\ $$$$\mathrm{V}_{\mathrm{n}} =\frac{\mathrm{n}^{\mathrm{P}} }{\mathrm{n}^{\mathrm{p}} }=\mathrm{1} \\ $$$$\mathrm{Lim}_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{U}_{\mathrm{n}} }{\mathrm{V}_{\mathrm{n}} }=\mathrm{lim}_{\mathrm{n}\rightarrow\infty\:\:} \frac{\mathrm{sin}\left(\mathrm{n}^{\mathrm{p}} \right)}{\mathrm{n}^{\mathrm{p}} }\:\:\:\mathrm{let}\:\mathrm{t}=\mathrm{n}^{\mathrm{p}} \\ $$$$=\mathrm{1} \\ $$$$ \\ $$$$\mathrm{P}=\mathrm{1} \\ $$$$\mathrm{Convergent}\:\mathrm{when}\:\mathrm{p}\geqslant\mathrm{1} \\ $$

Commented by mathmax by abdo last updated on 27/Mar/20

1) Σ_(n=1) ^∞ ((sin(n))/n) converges due to abel dirichlet theorem u_n =(1/n) is deceasing to 0 and ∃m>0 / ∣Σ_(k=0) ^(n ) sink∣<m

$$\left.\mathrm{1}\right)\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({n}\right)}{{n}}\:{converges}\:{due}\:{to}\:{abel}\:{dirichlet}\:{theorem} \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{{n}}\:{is}\:{deceasing}\:{to}\:\mathrm{0}\:\:{and}\:\exists{m}>\mathrm{0}\:\:/\:\mid\sum_{{k}=\mathrm{0}} ^{{n}\:} {sink}\mid<{m} \\ $$

Commented by mathmax by abdo last updated on 27/Mar/20

∣Σ_(n=1) ^∞ ((sin(n^p ))/n^p )∣≤Σ_(n=1) ^∞ (1/n^p ) so if p>1 this serie converges...

$$\mid\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({n}^{{p}} \right)}{{n}^{{p}} }\mid\leqslant\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{{p}} }\:\:{so}\:{if}\:{p}>\mathrm{1}\:\:{this}\:{serie}\:{converges}... \\ $$

Commented by Serlea last updated on 27/Mar/20

I think you are wrong bro

$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{are}\:\mathrm{wrong}\:\mathrm{bro} \\ $$

Commented by mathmax by abdo last updated on 27/Mar/20

i am not wrong revise courses of series....

$${i}\:{am}\:{not}\:{wrong}\:\:{revise}\:{courses}\:{of}\:{series}.... \\ $$

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