Does $sum_{n=1}^{infty} frac{1}{sqrt{n}-frac{2}3}$ converge or diverge?












2












$begingroup$


Does this series converge or diverge?



$$sum_{n=1}^{infty} frac{1}{sqrt{n}-frac{2}3}$$



I tried using the limit comparison test with $frac{1}{sqrt{n}}$, which diverges.



$$lim_{ntoinfty}{frac{{sqrt{n}}}{sqrt{n}-frac{2}3}}=1$$



Then the series diverges, is this right or I'm wrong?






sequences-and-series proof-verification convergence







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$endgroup$












  • $begingroup$
    looks good to me
    $endgroup$
    – gt6989b
    Dec 5 '18 at 20:19
















2












$begingroup$


Does this series converge or diverge?



$$sum_{n=1}^{infty} frac{1}{sqrt{n}-frac{2}3}$$



I tried using the limit comparison test with $frac{1}{sqrt{n}}$, which diverges.



$$lim_{ntoinfty}{frac{{sqrt{n}}}{sqrt{n}-frac{2}3}}=1$$



Then the series diverges, is this right or I'm wrong?






sequences-and-series proof-verification convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    looks good to me
    $endgroup$
    – gt6989b
    Dec 5 '18 at 20:19














2












2








2





$begingroup$


Does this series converge or diverge?



$$sum_{n=1}^{infty} frac{1}{sqrt{n}-frac{2}3}$$



I tried using the limit comparison test with $frac{1}{sqrt{n}}$, which diverges.



$$lim_{ntoinfty}{frac{{sqrt{n}}}{sqrt{n}-frac{2}3}}=1$$



Then the series diverges, is this right or I'm wrong?






sequences-and-series proof-verification convergence







share|cite|improve this question











$endgroup$




Does this series converge or diverge?



$$sum_{n=1}^{infty} frac{1}{sqrt{n}-frac{2}3}$$



I tried using the limit comparison test with $frac{1}{sqrt{n}}$, which diverges.



$$lim_{ntoinfty}{frac{{sqrt{n}}}{sqrt{n}-frac{2}3}}=1$$



Then the series diverges, is this right or I'm wrong?






sequences-and-series proof-verification convergence




sequences-and-series proof-verification convergence






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share|cite|improve this question













share|cite|improve this question




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edited Dec 22 '18 at 8:38









choco_addicted

8,07261847




8,07261847










asked Dec 5 '18 at 20:18









iggykimiiggykimi

1389




1389












  • $begingroup$
    looks good to me
    $endgroup$
    – gt6989b
    Dec 5 '18 at 20:19


















  • $begingroup$
    looks good to me
    $endgroup$
    – gt6989b
    Dec 5 '18 at 20:19
















$begingroup$
looks good to me
$endgroup$
– gt6989b
Dec 5 '18 at 20:19




$begingroup$
looks good to me
$endgroup$
– gt6989b
Dec 5 '18 at 20:19










2 Answers
2






active

oldest

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3












$begingroup$

Yes it is absolutely right, indeed note that



$$dfrac{1}{sqrt{n}-frac{2}3}sim dfrac{1}{sqrt{n}}$$



and the latter diverges for p test.



As an alternative by direct comparison test



$$dfrac{1}{sqrt{n}-frac{2}3}ge dfrac{1}{sqrt{n}}$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Yeah, just as another answer shows, you can see the convergence of such a sum of sequence does not matter w.r.t first several terms.



    Then, you can see



    $frac{1}{sqrt{n} -frac{2}{3}} sim frac{1}{sqrt{n}} = frac{1}{n^{1/2}}$



    notice that for those power less than or equals 1,(here it's 1/2), it's a diverge sequence.



    (You may refer to any analysis book for this result.)






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Yes it is absolutely right, indeed note that



      $$dfrac{1}{sqrt{n}-frac{2}3}sim dfrac{1}{sqrt{n}}$$



      and the latter diverges for p test.



      As an alternative by direct comparison test



      $$dfrac{1}{sqrt{n}-frac{2}3}ge dfrac{1}{sqrt{n}}$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Yes it is absolutely right, indeed note that



        $$dfrac{1}{sqrt{n}-frac{2}3}sim dfrac{1}{sqrt{n}}$$



        and the latter diverges for p test.



        As an alternative by direct comparison test



        $$dfrac{1}{sqrt{n}-frac{2}3}ge dfrac{1}{sqrt{n}}$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Yes it is absolutely right, indeed note that



          $$dfrac{1}{sqrt{n}-frac{2}3}sim dfrac{1}{sqrt{n}}$$



          and the latter diverges for p test.



          As an alternative by direct comparison test



          $$dfrac{1}{sqrt{n}-frac{2}3}ge dfrac{1}{sqrt{n}}$$






          share|cite|improve this answer









          $endgroup$



          Yes it is absolutely right, indeed note that



          $$dfrac{1}{sqrt{n}-frac{2}3}sim dfrac{1}{sqrt{n}}$$



          and the latter diverges for p test.



          As an alternative by direct comparison test



          $$dfrac{1}{sqrt{n}-frac{2}3}ge dfrac{1}{sqrt{n}}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 20:19









          gimusigimusi

          92.8k84494




          92.8k84494























              1












              $begingroup$

              Yeah, just as another answer shows, you can see the convergence of such a sum of sequence does not matter w.r.t first several terms.



              Then, you can see



              $frac{1}{sqrt{n} -frac{2}{3}} sim frac{1}{sqrt{n}} = frac{1}{n^{1/2}}$



              notice that for those power less than or equals 1,(here it's 1/2), it's a diverge sequence.



              (You may refer to any analysis book for this result.)






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Yeah, just as another answer shows, you can see the convergence of such a sum of sequence does not matter w.r.t first several terms.



                Then, you can see



                $frac{1}{sqrt{n} -frac{2}{3}} sim frac{1}{sqrt{n}} = frac{1}{n^{1/2}}$



                notice that for those power less than or equals 1,(here it's 1/2), it's a diverge sequence.



                (You may refer to any analysis book for this result.)






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Yeah, just as another answer shows, you can see the convergence of such a sum of sequence does not matter w.r.t first several terms.



                  Then, you can see



                  $frac{1}{sqrt{n} -frac{2}{3}} sim frac{1}{sqrt{n}} = frac{1}{n^{1/2}}$



                  notice that for those power less than or equals 1,(here it's 1/2), it's a diverge sequence.



                  (You may refer to any analysis book for this result.)






                  share|cite|improve this answer









                  $endgroup$



                  Yeah, just as another answer shows, you can see the convergence of such a sum of sequence does not matter w.r.t first several terms.



                  Then, you can see



                  $frac{1}{sqrt{n} -frac{2}{3}} sim frac{1}{sqrt{n}} = frac{1}{n^{1/2}}$



                  notice that for those power less than or equals 1,(here it's 1/2), it's a diverge sequence.



                  (You may refer to any analysis book for this result.)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 5 '18 at 20:24









                  汪铈达汪铈达

                  111




                  111






























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