Prove divergence of $sumlimits_{n=1}^inftyleft(frac1{sqrt n}-frac1{sqrt{n+1}}right)^{2/3}$












0












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Prove divergence of $sumlimits_{n=1}^inftyleft(frac1{sqrt n}-frac1{sqrt{n+1}}right)^{2/3}$.




I need to prove that this series is divergent, but the ratio test, root test and the divergent majorant $frac{1}{n}$ are inconclusive, so how would I be able to prove this?



I also tried it being proportional to something, but I couldn't get a divergent majorant out of it, because for large n it always becomes smaller.






real-analysis sequences-and-series convergence







share|cite|improve this question











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  • 4




    $begingroup$
    $$left(frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}}right)^{2/3}=left(frac{1}{sqrt{n},sqrt{n+1},(sqrt{n+1}+sqrt n)}right)^{2/3}simleft(frac{1}{2n^{3/2}}right)^{2/3}=cdots$$
    $endgroup$
    – Did
    Jan 23 at 9:33












  • $begingroup$
    @Did In your opinion is "This question is missing context or other details"?
    $endgroup$
    – Robert Z
    Jan 23 at 9:47










  • $begingroup$
    Related: For what values $alpha$ does the sum converge?.
    $endgroup$
    – Martin R
    Jan 23 at 9:50










  • $begingroup$
    @RobertZ It is borderline, on the one hand the OP mentions some tries, on the other hand some declarations are suspect, for example " I couldn't get a divergent majorant out of it, because for large n it always becomes smaller". Why do you ask?
    $endgroup$
    – Did
    Jan 23 at 9:53










  • $begingroup$
    @Did Because a few weeks ago I answered to a very similar "borderline" question and then the question has been quickly deleted. I wonder what is going happen to this question...
    $endgroup$
    – Robert Z
    Jan 23 at 10:00
















0












$begingroup$



Prove divergence of $sumlimits_{n=1}^inftyleft(frac1{sqrt n}-frac1{sqrt{n+1}}right)^{2/3}$.




I need to prove that this series is divergent, but the ratio test, root test and the divergent majorant $frac{1}{n}$ are inconclusive, so how would I be able to prove this?



I also tried it being proportional to something, but I couldn't get a divergent majorant out of it, because for large n it always becomes smaller.






real-analysis sequences-and-series convergence







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    $$left(frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}}right)^{2/3}=left(frac{1}{sqrt{n},sqrt{n+1},(sqrt{n+1}+sqrt n)}right)^{2/3}simleft(frac{1}{2n^{3/2}}right)^{2/3}=cdots$$
    $endgroup$
    – Did
    Jan 23 at 9:33












  • $begingroup$
    @Did In your opinion is "This question is missing context or other details"?
    $endgroup$
    – Robert Z
    Jan 23 at 9:47










  • $begingroup$
    Related: For what values $alpha$ does the sum converge?.
    $endgroup$
    – Martin R
    Jan 23 at 9:50










  • $begingroup$
    @RobertZ It is borderline, on the one hand the OP mentions some tries, on the other hand some declarations are suspect, for example " I couldn't get a divergent majorant out of it, because for large n it always becomes smaller". Why do you ask?
    $endgroup$
    – Did
    Jan 23 at 9:53










  • $begingroup$
    @Did Because a few weeks ago I answered to a very similar "borderline" question and then the question has been quickly deleted. I wonder what is going happen to this question...
    $endgroup$
    – Robert Z
    Jan 23 at 10:00














0












0








0





$begingroup$



Prove divergence of $sumlimits_{n=1}^inftyleft(frac1{sqrt n}-frac1{sqrt{n+1}}right)^{2/3}$.




I need to prove that this series is divergent, but the ratio test, root test and the divergent majorant $frac{1}{n}$ are inconclusive, so how would I be able to prove this?



I also tried it being proportional to something, but I couldn't get a divergent majorant out of it, because for large n it always becomes smaller.






real-analysis sequences-and-series convergence







share|cite|improve this question











$endgroup$





Prove divergence of $sumlimits_{n=1}^inftyleft(frac1{sqrt n}-frac1{sqrt{n+1}}right)^{2/3}$.




I need to prove that this series is divergent, but the ratio test, root test and the divergent majorant $frac{1}{n}$ are inconclusive, so how would I be able to prove this?



I also tried it being proportional to something, but I couldn't get a divergent majorant out of it, because for large n it always becomes smaller.






real-analysis sequences-and-series convergence




real-analysis sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 9:36









Did

248k23225463




248k23225463










asked Jan 23 at 9:29









C. EliasC. Elias

223




223








  • 4




    $begingroup$
    $$left(frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}}right)^{2/3}=left(frac{1}{sqrt{n},sqrt{n+1},(sqrt{n+1}+sqrt n)}right)^{2/3}simleft(frac{1}{2n^{3/2}}right)^{2/3}=cdots$$
    $endgroup$
    – Did
    Jan 23 at 9:33












  • $begingroup$
    @Did In your opinion is "This question is missing context or other details"?
    $endgroup$
    – Robert Z
    Jan 23 at 9:47










  • $begingroup$
    Related: For what values $alpha$ does the sum converge?.
    $endgroup$
    – Martin R
    Jan 23 at 9:50










  • $begingroup$
    @RobertZ It is borderline, on the one hand the OP mentions some tries, on the other hand some declarations are suspect, for example " I couldn't get a divergent majorant out of it, because for large n it always becomes smaller". Why do you ask?
    $endgroup$
    – Did
    Jan 23 at 9:53










  • $begingroup$
    @Did Because a few weeks ago I answered to a very similar "borderline" question and then the question has been quickly deleted. I wonder what is going happen to this question...
    $endgroup$
    – Robert Z
    Jan 23 at 10:00














  • 4




    $begingroup$
    $$left(frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}}right)^{2/3}=left(frac{1}{sqrt{n},sqrt{n+1},(sqrt{n+1}+sqrt n)}right)^{2/3}simleft(frac{1}{2n^{3/2}}right)^{2/3}=cdots$$
    $endgroup$
    – Did
    Jan 23 at 9:33












  • $begingroup$
    @Did In your opinion is "This question is missing context or other details"?
    $endgroup$
    – Robert Z
    Jan 23 at 9:47










  • $begingroup$
    Related: For what values $alpha$ does the sum converge?.
    $endgroup$
    – Martin R
    Jan 23 at 9:50










  • $begingroup$
    @RobertZ It is borderline, on the one hand the OP mentions some tries, on the other hand some declarations are suspect, for example " I couldn't get a divergent majorant out of it, because for large n it always becomes smaller". Why do you ask?
    $endgroup$
    – Did
    Jan 23 at 9:53










  • $begingroup$
    @Did Because a few weeks ago I answered to a very similar "borderline" question and then the question has been quickly deleted. I wonder what is going happen to this question...
    $endgroup$
    – Robert Z
    Jan 23 at 10:00








4




4




$begingroup$
$$left(frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}}right)^{2/3}=left(frac{1}{sqrt{n},sqrt{n+1},(sqrt{n+1}+sqrt n)}right)^{2/3}simleft(frac{1}{2n^{3/2}}right)^{2/3}=cdots$$
$endgroup$
– Did
Jan 23 at 9:33






$begingroup$
$$left(frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}}right)^{2/3}=left(frac{1}{sqrt{n},sqrt{n+1},(sqrt{n+1}+sqrt n)}right)^{2/3}simleft(frac{1}{2n^{3/2}}right)^{2/3}=cdots$$
$endgroup$
– Did
Jan 23 at 9:33














$begingroup$
@Did In your opinion is "This question is missing context or other details"?
$endgroup$
– Robert Z
Jan 23 at 9:47




$begingroup$
@Did In your opinion is "This question is missing context or other details"?
$endgroup$
– Robert Z
Jan 23 at 9:47












$begingroup$
Related: For what values $alpha$ does the sum converge?.
$endgroup$
– Martin R
Jan 23 at 9:50




$begingroup$
Related: For what values $alpha$ does the sum converge?.
$endgroup$
– Martin R
Jan 23 at 9:50












$begingroup$
@RobertZ It is borderline, on the one hand the OP mentions some tries, on the other hand some declarations are suspect, for example " I couldn't get a divergent majorant out of it, because for large n it always becomes smaller". Why do you ask?
$endgroup$
– Did
Jan 23 at 9:53




$begingroup$
@RobertZ It is borderline, on the one hand the OP mentions some tries, on the other hand some declarations are suspect, for example " I couldn't get a divergent majorant out of it, because for large n it always becomes smaller". Why do you ask?
$endgroup$
– Did
Jan 23 at 9:53












$begingroup$
@Did Because a few weeks ago I answered to a very similar "borderline" question and then the question has been quickly deleted. I wonder what is going happen to this question...
$endgroup$
– Robert Z
Jan 23 at 10:00




$begingroup$
@Did Because a few weeks ago I answered to a very similar "borderline" question and then the question has been quickly deleted. I wonder what is going happen to this question...
$endgroup$
– Robert Z
Jan 23 at 10:00










3 Answers
3






active

oldest

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3












$begingroup$

Note that
$$
frac 1{sqrt n} - frac 1{sqrt {n+1}} = frac {sqrt{n+1} - sqrt n}{sqrt {(n+1)n}} = frac 1 {sqrt {n(n+1)} cdot (sqrt n + sqrt{n+1})} sim frac 1{2n^{3/2}} quad [n to infty]
$$
hence the term is asymptotically $(1/2)^{2/3} n^{-1}$, and the series diverges.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $frac 1 {sqrt n} -frac 1 {sqrt {n+1}} =frac 1 {sqrt n sqrt {n+1}(sqrt n +sqrt {n+1})}$. Compare the given series with $sum frac 1 n$ which is divergent.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      $$left(frac1{sqrt n}-frac1{sqrt{n+1}}right)^{2/3}=left(frac{sqrt{n+1}-sqrt n}{sqrt nsqrt{n+1}}right)^{2/3}=left(frac{1}{sqrt nsqrt{n+1}(sqrt{n+1}+sqrt n)}right)^{2/3}>left(frac{1}{sqrt{n+1}sqrt{n+1}(sqrt{n+1}+sqrt{n+1})}right)^{2/3}=frac{2^{-2/3}}{n+1}$$



      so that the majorant test is conclusive.






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Note that
        $$
        frac 1{sqrt n} - frac 1{sqrt {n+1}} = frac {sqrt{n+1} - sqrt n}{sqrt {(n+1)n}} = frac 1 {sqrt {n(n+1)} cdot (sqrt n + sqrt{n+1})} sim frac 1{2n^{3/2}} quad [n to infty]
        $$
        hence the term is asymptotically $(1/2)^{2/3} n^{-1}$, and the series diverges.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Note that
          $$
          frac 1{sqrt n} - frac 1{sqrt {n+1}} = frac {sqrt{n+1} - sqrt n}{sqrt {(n+1)n}} = frac 1 {sqrt {n(n+1)} cdot (sqrt n + sqrt{n+1})} sim frac 1{2n^{3/2}} quad [n to infty]
          $$
          hence the term is asymptotically $(1/2)^{2/3} n^{-1}$, and the series diverges.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Note that
            $$
            frac 1{sqrt n} - frac 1{sqrt {n+1}} = frac {sqrt{n+1} - sqrt n}{sqrt {(n+1)n}} = frac 1 {sqrt {n(n+1)} cdot (sqrt n + sqrt{n+1})} sim frac 1{2n^{3/2}} quad [n to infty]
            $$
            hence the term is asymptotically $(1/2)^{2/3} n^{-1}$, and the series diverges.






            share|cite|improve this answer









            $endgroup$



            Note that
            $$
            frac 1{sqrt n} - frac 1{sqrt {n+1}} = frac {sqrt{n+1} - sqrt n}{sqrt {(n+1)n}} = frac 1 {sqrt {n(n+1)} cdot (sqrt n + sqrt{n+1})} sim frac 1{2n^{3/2}} quad [n to infty]
            $$
            hence the term is asymptotically $(1/2)^{2/3} n^{-1}$, and the series diverges.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 23 at 9:35









            xbhxbh

            6,3001522




            6,3001522























                0












                $begingroup$

                $frac 1 {sqrt n} -frac 1 {sqrt {n+1}} =frac 1 {sqrt n sqrt {n+1}(sqrt n +sqrt {n+1})}$. Compare the given series with $sum frac 1 n$ which is divergent.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  $frac 1 {sqrt n} -frac 1 {sqrt {n+1}} =frac 1 {sqrt n sqrt {n+1}(sqrt n +sqrt {n+1})}$. Compare the given series with $sum frac 1 n$ which is divergent.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    $frac 1 {sqrt n} -frac 1 {sqrt {n+1}} =frac 1 {sqrt n sqrt {n+1}(sqrt n +sqrt {n+1})}$. Compare the given series with $sum frac 1 n$ which is divergent.






                    share|cite|improve this answer









                    $endgroup$



                    $frac 1 {sqrt n} -frac 1 {sqrt {n+1}} =frac 1 {sqrt n sqrt {n+1}(sqrt n +sqrt {n+1})}$. Compare the given series with $sum frac 1 n$ which is divergent.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 23 at 9:33









                    Kavi Rama MurthyKavi Rama Murthy

                    66.8k53067




                    66.8k53067























                        0












                        $begingroup$

                        $$left(frac1{sqrt n}-frac1{sqrt{n+1}}right)^{2/3}=left(frac{sqrt{n+1}-sqrt n}{sqrt nsqrt{n+1}}right)^{2/3}=left(frac{1}{sqrt nsqrt{n+1}(sqrt{n+1}+sqrt n)}right)^{2/3}>left(frac{1}{sqrt{n+1}sqrt{n+1}(sqrt{n+1}+sqrt{n+1})}right)^{2/3}=frac{2^{-2/3}}{n+1}$$



                        so that the majorant test is conclusive.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          $$left(frac1{sqrt n}-frac1{sqrt{n+1}}right)^{2/3}=left(frac{sqrt{n+1}-sqrt n}{sqrt nsqrt{n+1}}right)^{2/3}=left(frac{1}{sqrt nsqrt{n+1}(sqrt{n+1}+sqrt n)}right)^{2/3}>left(frac{1}{sqrt{n+1}sqrt{n+1}(sqrt{n+1}+sqrt{n+1})}right)^{2/3}=frac{2^{-2/3}}{n+1}$$



                          so that the majorant test is conclusive.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            $$left(frac1{sqrt n}-frac1{sqrt{n+1}}right)^{2/3}=left(frac{sqrt{n+1}-sqrt n}{sqrt nsqrt{n+1}}right)^{2/3}=left(frac{1}{sqrt nsqrt{n+1}(sqrt{n+1}+sqrt n)}right)^{2/3}>left(frac{1}{sqrt{n+1}sqrt{n+1}(sqrt{n+1}+sqrt{n+1})}right)^{2/3}=frac{2^{-2/3}}{n+1}$$



                            so that the majorant test is conclusive.






                            share|cite|improve this answer









                            $endgroup$



                            $$left(frac1{sqrt n}-frac1{sqrt{n+1}}right)^{2/3}=left(frac{sqrt{n+1}-sqrt n}{sqrt nsqrt{n+1}}right)^{2/3}=left(frac{1}{sqrt nsqrt{n+1}(sqrt{n+1}+sqrt n)}right)^{2/3}>left(frac{1}{sqrt{n+1}sqrt{n+1}(sqrt{n+1}+sqrt{n+1})}right)^{2/3}=frac{2^{-2/3}}{n+1}$$



                            so that the majorant test is conclusive.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 23 at 9:45









                            Yves DaoustYves Daoust

                            130k676227




                            130k676227






























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