Why doesn't $sum frac{sin(frac{1}{n})}{sqrt(n)}$ diverge?












2












$begingroup$


Why doesn't $sum frac{sin(frac{1}{n})}{sqrt(n)}$ diverge?



I know that it converges, I just want to know what i'm doing wrong here. Here's what I did.



$sumfrac{-1}{sqrt n} < $ $sum frac{sin(frac{1}{n})}{sqrt n}$ $sum<frac{-1}{sqrt n }$.



Both $sumfrac{-1}{sqrt n}$ and $sum frac{-1}{sqrt n} $ diverge by the p series test since p=0.5<1.
Therefore $sum frac{sin(frac{1}{n})}{sqrt n}$ also diverges. What am I doing wrong here?






sequences-and-series convergence divergent-series







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  • 1




    $begingroup$
    comparison-test is valid for positive series
    $endgroup$
    – Minz
    Dec 8 '18 at 2:59






  • 1




    $begingroup$
    $dfrac{sinleft(dfrac1nright)}{sqrt{n}}simdfrac{1}{nsqrt{n}}$ (asymptotic) which converges by p series test.
    $endgroup$
    – Yadati Kiran
    Dec 8 '18 at 2:59


















2












$begingroup$


Why doesn't $sum frac{sin(frac{1}{n})}{sqrt(n)}$ diverge?



I know that it converges, I just want to know what i'm doing wrong here. Here's what I did.



$sumfrac{-1}{sqrt n} < $ $sum frac{sin(frac{1}{n})}{sqrt n}$ $sum<frac{-1}{sqrt n }$.



Both $sumfrac{-1}{sqrt n}$ and $sum frac{-1}{sqrt n} $ diverge by the p series test since p=0.5<1.
Therefore $sum frac{sin(frac{1}{n})}{sqrt n}$ also diverges. What am I doing wrong here?






sequences-and-series convergence divergent-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    comparison-test is valid for positive series
    $endgroup$
    – Minz
    Dec 8 '18 at 2:59






  • 1




    $begingroup$
    $dfrac{sinleft(dfrac1nright)}{sqrt{n}}simdfrac{1}{nsqrt{n}}$ (asymptotic) which converges by p series test.
    $endgroup$
    – Yadati Kiran
    Dec 8 '18 at 2:59
















2












2








2





$begingroup$


Why doesn't $sum frac{sin(frac{1}{n})}{sqrt(n)}$ diverge?



I know that it converges, I just want to know what i'm doing wrong here. Here's what I did.



$sumfrac{-1}{sqrt n} < $ $sum frac{sin(frac{1}{n})}{sqrt n}$ $sum<frac{-1}{sqrt n }$.



Both $sumfrac{-1}{sqrt n}$ and $sum frac{-1}{sqrt n} $ diverge by the p series test since p=0.5<1.
Therefore $sum frac{sin(frac{1}{n})}{sqrt n}$ also diverges. What am I doing wrong here?






sequences-and-series convergence divergent-series







share|cite|improve this question









$endgroup$




Why doesn't $sum frac{sin(frac{1}{n})}{sqrt(n)}$ diverge?



I know that it converges, I just want to know what i'm doing wrong here. Here's what I did.



$sumfrac{-1}{sqrt n} < $ $sum frac{sin(frac{1}{n})}{sqrt n}$ $sum<frac{-1}{sqrt n }$.



Both $sumfrac{-1}{sqrt n}$ and $sum frac{-1}{sqrt n} $ diverge by the p series test since p=0.5<1.
Therefore $sum frac{sin(frac{1}{n})}{sqrt n}$ also diverges. What am I doing wrong here?






sequences-and-series convergence divergent-series




sequences-and-series convergence divergent-series






share|cite|improve this question













share|cite|improve this question











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asked Dec 8 '18 at 2:53









user140161user140161

664617




664617








  • 1




    $begingroup$
    comparison-test is valid for positive series
    $endgroup$
    – Minz
    Dec 8 '18 at 2:59






  • 1




    $begingroup$
    $dfrac{sinleft(dfrac1nright)}{sqrt{n}}simdfrac{1}{nsqrt{n}}$ (asymptotic) which converges by p series test.
    $endgroup$
    – Yadati Kiran
    Dec 8 '18 at 2:59
















  • 1




    $begingroup$
    comparison-test is valid for positive series
    $endgroup$
    – Minz
    Dec 8 '18 at 2:59






  • 1




    $begingroup$
    $dfrac{sinleft(dfrac1nright)}{sqrt{n}}simdfrac{1}{nsqrt{n}}$ (asymptotic) which converges by p series test.
    $endgroup$
    – Yadati Kiran
    Dec 8 '18 at 2:59










1




1




$begingroup$
comparison-test is valid for positive series
$endgroup$
– Minz
Dec 8 '18 at 2:59




$begingroup$
comparison-test is valid for positive series
$endgroup$
– Minz
Dec 8 '18 at 2:59




1




1




$begingroup$
$dfrac{sinleft(dfrac1nright)}{sqrt{n}}simdfrac{1}{nsqrt{n}}$ (asymptotic) which converges by p series test.
$endgroup$
– Yadati Kiran
Dec 8 '18 at 2:59






$begingroup$
$dfrac{sinleft(dfrac1nright)}{sqrt{n}}simdfrac{1}{nsqrt{n}}$ (asymptotic) which converges by p series test.
$endgroup$
– Yadati Kiran
Dec 8 '18 at 2:59












2 Answers
2






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4












$begingroup$

Your implication is wrong. For example, we know that,$$n^2>sqrt n forall ninmathbb N$$
$$Longrightarrowfrac{1}{n^2}<frac{1}{sqrt n} forall ninmathbb N$$
$$Longrightarrowfrac{-1}{sqrt n}<frac{1}{n^2}<frac{1}{sqrt n} forall ninmathbb N$$
$$Longrightarrowsumfrac{-1}{sqrt n}<sumfrac{1}{n^2}<sumfrac{1}{sqrt n} forall ninmathbb N$$
But we know that $sumfrac{1}{n^2}$ converges, whereas others diverge.



Hope it is helpful






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    For $n$ large, $sin (1/n) > 0$ and the terms are positive ,and using $sin (1/n) < 1/n $ we have the given series converges by comparing it with a $p =:3/2$ series.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Your implication is wrong. For example, we know that,$$n^2>sqrt n forall ninmathbb N$$
      $$Longrightarrowfrac{1}{n^2}<frac{1}{sqrt n} forall ninmathbb N$$
      $$Longrightarrowfrac{-1}{sqrt n}<frac{1}{n^2}<frac{1}{sqrt n} forall ninmathbb N$$
      $$Longrightarrowsumfrac{-1}{sqrt n}<sumfrac{1}{n^2}<sumfrac{1}{sqrt n} forall ninmathbb N$$
      But we know that $sumfrac{1}{n^2}$ converges, whereas others diverge.



      Hope it is helpful






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Your implication is wrong. For example, we know that,$$n^2>sqrt n forall ninmathbb N$$
        $$Longrightarrowfrac{1}{n^2}<frac{1}{sqrt n} forall ninmathbb N$$
        $$Longrightarrowfrac{-1}{sqrt n}<frac{1}{n^2}<frac{1}{sqrt n} forall ninmathbb N$$
        $$Longrightarrowsumfrac{-1}{sqrt n}<sumfrac{1}{n^2}<sumfrac{1}{sqrt n} forall ninmathbb N$$
        But we know that $sumfrac{1}{n^2}$ converges, whereas others diverge.



        Hope it is helpful






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Your implication is wrong. For example, we know that,$$n^2>sqrt n forall ninmathbb N$$
          $$Longrightarrowfrac{1}{n^2}<frac{1}{sqrt n} forall ninmathbb N$$
          $$Longrightarrowfrac{-1}{sqrt n}<frac{1}{n^2}<frac{1}{sqrt n} forall ninmathbb N$$
          $$Longrightarrowsumfrac{-1}{sqrt n}<sumfrac{1}{n^2}<sumfrac{1}{sqrt n} forall ninmathbb N$$
          But we know that $sumfrac{1}{n^2}$ converges, whereas others diverge.



          Hope it is helpful






          share|cite|improve this answer









          $endgroup$



          Your implication is wrong. For example, we know that,$$n^2>sqrt n forall ninmathbb N$$
          $$Longrightarrowfrac{1}{n^2}<frac{1}{sqrt n} forall ninmathbb N$$
          $$Longrightarrowfrac{-1}{sqrt n}<frac{1}{n^2}<frac{1}{sqrt n} forall ninmathbb N$$
          $$Longrightarrowsumfrac{-1}{sqrt n}<sumfrac{1}{n^2}<sumfrac{1}{sqrt n} forall ninmathbb N$$
          But we know that $sumfrac{1}{n^2}$ converges, whereas others diverge.



          Hope it is helpful







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 3:07









          MartundMartund

          1,665213




          1,665213























              0












              $begingroup$

              For $n$ large, $sin (1/n) > 0$ and the terms are positive ,and using $sin (1/n) < 1/n $ we have the given series converges by comparing it with a $p =:3/2$ series.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                For $n$ large, $sin (1/n) > 0$ and the terms are positive ,and using $sin (1/n) < 1/n $ we have the given series converges by comparing it with a $p =:3/2$ series.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  For $n$ large, $sin (1/n) > 0$ and the terms are positive ,and using $sin (1/n) < 1/n $ we have the given series converges by comparing it with a $p =:3/2$ series.






                  share|cite|improve this answer









                  $endgroup$



                  For $n$ large, $sin (1/n) > 0$ and the terms are positive ,and using $sin (1/n) < 1/n $ we have the given series converges by comparing it with a $p =:3/2$ series.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 8 '18 at 3:00









                  DeepSeaDeepSea

                  71.3k54488




                  71.3k54488






























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