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1-Determine-the-following-if-it-is-convergent-or-divergent-n-1-sin-n-n-2-n-1-sin-n-p-n-p-p-R-find-the-range-of-p-when-it-is-convergent-




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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