Determine all $q in mathbb{Q}$, so that $ sum_{n=1}^{infty}{frac{sqrt{n+1}-sqrt{n}}{n^q}} $ converges












0














I already tried the ratio and root criterion, but it didn't get me anywhere. I'd be thrilled if you had any suggestions.



(also applied the third binomial formula, so it would "look nicer", maybe it helps $sum_{n=1}^{infty}{frac{sqrt{n+1}-sqrt{n}}{n^q}}=sum_{n=1}^{infty}{frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$)



thanks and greetings!






real-analysis sequences-and-series convergence rational-numbers







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    0














    I already tried the ratio and root criterion, but it didn't get me anywhere. I'd be thrilled if you had any suggestions.



    (also applied the third binomial formula, so it would "look nicer", maybe it helps $sum_{n=1}^{infty}{frac{sqrt{n+1}-sqrt{n}}{n^q}}=sum_{n=1}^{infty}{frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$)



    thanks and greetings!






    real-analysis sequences-and-series convergence rational-numbers







    share|cite|improve this question

























      0












      0








      0


      1





      I already tried the ratio and root criterion, but it didn't get me anywhere. I'd be thrilled if you had any suggestions.



      (also applied the third binomial formula, so it would "look nicer", maybe it helps $sum_{n=1}^{infty}{frac{sqrt{n+1}-sqrt{n}}{n^q}}=sum_{n=1}^{infty}{frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$)



      thanks and greetings!






      real-analysis sequences-and-series convergence rational-numbers







      share|cite|improve this question













      I already tried the ratio and root criterion, but it didn't get me anywhere. I'd be thrilled if you had any suggestions.



      (also applied the third binomial formula, so it would "look nicer", maybe it helps $sum_{n=1}^{infty}{frac{sqrt{n+1}-sqrt{n}}{n^q}}=sum_{n=1}^{infty}{frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$)



      thanks and greetings!






      real-analysis sequences-and-series convergence rational-numbers




      real-analysis sequences-and-series convergence rational-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 29 '18 at 15:28









      DDevelopsDDevelops

      533




      533






















          3 Answers
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          Clearly
          $$
          frac{1}{2(n+1)^{q+{1/2}}}<frac{sqrt{n+1}-sqrt{n}}{n^q}=frac{1}{(sqrt{n+1}+sqrt{n})cdot n^q}<frac{1}{2n^{q+{1/2}}},
          $$
          and $sum_{n=1}^inftyfrac{1}{n^{q+1/2}}$, equivalently $sum_{n=1}^inftyfrac{1}{(n+1)^{q+1/2}}$, converge, if and only if $q+1/2>1$, equivalently $q>1/2$.






          share|cite|improve this answer























          • Equivalently $q>-1/2$?
            – gimusi
            Nov 29 '18 at 15:47










          • Thanks! I corrected it!
            – Yiorgos S. Smyrlis
            Nov 29 '18 at 15:49










          • You are welcome! minor typo of course!
            – gimusi
            Nov 29 '18 at 15:50










          • thx a lot, that helps, I assume you conclude the convergence by the direct comparison test. Another tiny "problem": You showed that for those q that suffice $q>-frac{1}{2}$ the series must converge. But how do we know there is no q, for which $frac{1}{2n^{q+0.5}}$ diverges but $frac{1}{sqrt{n+1}+sqrt{n}}$ still converges. Do I have to show that or is it obvious/trivial?
            – DDevelops
            Nov 29 '18 at 17:05





















          1














          Hint 1:



          $${frac{sqrt{n+1}-sqrt{n}}{n^q}}={frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$$



          Hint 2 Limit compare ${frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$ to ${frac{1}{n^qsqrt{n}}}$






          share|cite|improve this answer





























            1














            From here



            $$sum_{n=1}^{infty}{frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$$



            we have that



            $$frac{1}{n^q(sqrt{n+1}+sqrt{n})}sim frac{1}{2n^{q+frac12}}$$



            then refer to limit comparison test to conlcude that $q+frac12>1$.






            share|cite|improve this answer























              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2














              Clearly
              $$
              frac{1}{2(n+1)^{q+{1/2}}}<frac{sqrt{n+1}-sqrt{n}}{n^q}=frac{1}{(sqrt{n+1}+sqrt{n})cdot n^q}<frac{1}{2n^{q+{1/2}}},
              $$
              and $sum_{n=1}^inftyfrac{1}{n^{q+1/2}}$, equivalently $sum_{n=1}^inftyfrac{1}{(n+1)^{q+1/2}}$, converge, if and only if $q+1/2>1$, equivalently $q>1/2$.






              share|cite|improve this answer























              • Equivalently $q>-1/2$?
                – gimusi
                Nov 29 '18 at 15:47










              • Thanks! I corrected it!
                – Yiorgos S. Smyrlis
                Nov 29 '18 at 15:49










              • You are welcome! minor typo of course!
                – gimusi
                Nov 29 '18 at 15:50










              • thx a lot, that helps, I assume you conclude the convergence by the direct comparison test. Another tiny "problem": You showed that for those q that suffice $q>-frac{1}{2}$ the series must converge. But how do we know there is no q, for which $frac{1}{2n^{q+0.5}}$ diverges but $frac{1}{sqrt{n+1}+sqrt{n}}$ still converges. Do I have to show that or is it obvious/trivial?
                – DDevelops
                Nov 29 '18 at 17:05


















              2














              Clearly
              $$
              frac{1}{2(n+1)^{q+{1/2}}}<frac{sqrt{n+1}-sqrt{n}}{n^q}=frac{1}{(sqrt{n+1}+sqrt{n})cdot n^q}<frac{1}{2n^{q+{1/2}}},
              $$
              and $sum_{n=1}^inftyfrac{1}{n^{q+1/2}}$, equivalently $sum_{n=1}^inftyfrac{1}{(n+1)^{q+1/2}}$, converge, if and only if $q+1/2>1$, equivalently $q>1/2$.






              share|cite|improve this answer























              • Equivalently $q>-1/2$?
                – gimusi
                Nov 29 '18 at 15:47










              • Thanks! I corrected it!
                – Yiorgos S. Smyrlis
                Nov 29 '18 at 15:49










              • You are welcome! minor typo of course!
                – gimusi
                Nov 29 '18 at 15:50










              • thx a lot, that helps, I assume you conclude the convergence by the direct comparison test. Another tiny "problem": You showed that for those q that suffice $q>-frac{1}{2}$ the series must converge. But how do we know there is no q, for which $frac{1}{2n^{q+0.5}}$ diverges but $frac{1}{sqrt{n+1}+sqrt{n}}$ still converges. Do I have to show that or is it obvious/trivial?
                – DDevelops
                Nov 29 '18 at 17:05
















              2












              2








              2






              Clearly
              $$
              frac{1}{2(n+1)^{q+{1/2}}}<frac{sqrt{n+1}-sqrt{n}}{n^q}=frac{1}{(sqrt{n+1}+sqrt{n})cdot n^q}<frac{1}{2n^{q+{1/2}}},
              $$
              and $sum_{n=1}^inftyfrac{1}{n^{q+1/2}}$, equivalently $sum_{n=1}^inftyfrac{1}{(n+1)^{q+1/2}}$, converge, if and only if $q+1/2>1$, equivalently $q>1/2$.






              share|cite|improve this answer














              Clearly
              $$
              frac{1}{2(n+1)^{q+{1/2}}}<frac{sqrt{n+1}-sqrt{n}}{n^q}=frac{1}{(sqrt{n+1}+sqrt{n})cdot n^q}<frac{1}{2n^{q+{1/2}}},
              $$
              and $sum_{n=1}^inftyfrac{1}{n^{q+1/2}}$, equivalently $sum_{n=1}^inftyfrac{1}{(n+1)^{q+1/2}}$, converge, if and only if $q+1/2>1$, equivalently $q>1/2$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 29 '18 at 15:49

























              answered Nov 29 '18 at 15:44









              Yiorgos S. SmyrlisYiorgos S. Smyrlis

              62.8k1383163




              62.8k1383163












              • Equivalently $q>-1/2$?
                – gimusi
                Nov 29 '18 at 15:47










              • Thanks! I corrected it!
                – Yiorgos S. Smyrlis
                Nov 29 '18 at 15:49










              • You are welcome! minor typo of course!
                – gimusi
                Nov 29 '18 at 15:50










              • thx a lot, that helps, I assume you conclude the convergence by the direct comparison test. Another tiny "problem": You showed that for those q that suffice $q>-frac{1}{2}$ the series must converge. But how do we know there is no q, for which $frac{1}{2n^{q+0.5}}$ diverges but $frac{1}{sqrt{n+1}+sqrt{n}}$ still converges. Do I have to show that or is it obvious/trivial?
                – DDevelops
                Nov 29 '18 at 17:05




















              • Equivalently $q>-1/2$?
                – gimusi
                Nov 29 '18 at 15:47










              • Thanks! I corrected it!
                – Yiorgos S. Smyrlis
                Nov 29 '18 at 15:49










              • You are welcome! minor typo of course!
                – gimusi
                Nov 29 '18 at 15:50










              • thx a lot, that helps, I assume you conclude the convergence by the direct comparison test. Another tiny "problem": You showed that for those q that suffice $q>-frac{1}{2}$ the series must converge. But how do we know there is no q, for which $frac{1}{2n^{q+0.5}}$ diverges but $frac{1}{sqrt{n+1}+sqrt{n}}$ still converges. Do I have to show that or is it obvious/trivial?
                – DDevelops
                Nov 29 '18 at 17:05


















              Equivalently $q>-1/2$?
              – gimusi
              Nov 29 '18 at 15:47




              Equivalently $q>-1/2$?
              – gimusi
              Nov 29 '18 at 15:47












              Thanks! I corrected it!
              – Yiorgos S. Smyrlis
              Nov 29 '18 at 15:49




              Thanks! I corrected it!
              – Yiorgos S. Smyrlis
              Nov 29 '18 at 15:49












              You are welcome! minor typo of course!
              – gimusi
              Nov 29 '18 at 15:50




              You are welcome! minor typo of course!
              – gimusi
              Nov 29 '18 at 15:50












              thx a lot, that helps, I assume you conclude the convergence by the direct comparison test. Another tiny "problem": You showed that for those q that suffice $q>-frac{1}{2}$ the series must converge. But how do we know there is no q, for which $frac{1}{2n^{q+0.5}}$ diverges but $frac{1}{sqrt{n+1}+sqrt{n}}$ still converges. Do I have to show that or is it obvious/trivial?
              – DDevelops
              Nov 29 '18 at 17:05






              thx a lot, that helps, I assume you conclude the convergence by the direct comparison test. Another tiny "problem": You showed that for those q that suffice $q>-frac{1}{2}$ the series must converge. But how do we know there is no q, for which $frac{1}{2n^{q+0.5}}$ diverges but $frac{1}{sqrt{n+1}+sqrt{n}}$ still converges. Do I have to show that or is it obvious/trivial?
              – DDevelops
              Nov 29 '18 at 17:05













              1














              Hint 1:



              $${frac{sqrt{n+1}-sqrt{n}}{n^q}}={frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$$



              Hint 2 Limit compare ${frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$ to ${frac{1}{n^qsqrt{n}}}$






              share|cite|improve this answer


























                1














                Hint 1:



                $${frac{sqrt{n+1}-sqrt{n}}{n^q}}={frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$$



                Hint 2 Limit compare ${frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$ to ${frac{1}{n^qsqrt{n}}}$






                share|cite|improve this answer
























                  1












                  1








                  1






                  Hint 1:



                  $${frac{sqrt{n+1}-sqrt{n}}{n^q}}={frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$$



                  Hint 2 Limit compare ${frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$ to ${frac{1}{n^qsqrt{n}}}$






                  share|cite|improve this answer












                  Hint 1:



                  $${frac{sqrt{n+1}-sqrt{n}}{n^q}}={frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$$



                  Hint 2 Limit compare ${frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$ to ${frac{1}{n^qsqrt{n}}}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 29 '18 at 15:33









                  N. S.N. S.

                  102k6111207




                  102k6111207























                      1














                      From here



                      $$sum_{n=1}^{infty}{frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$$



                      we have that



                      $$frac{1}{n^q(sqrt{n+1}+sqrt{n})}sim frac{1}{2n^{q+frac12}}$$



                      then refer to limit comparison test to conlcude that $q+frac12>1$.






                      share|cite|improve this answer




























                        1














                        From here



                        $$sum_{n=1}^{infty}{frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$$



                        we have that



                        $$frac{1}{n^q(sqrt{n+1}+sqrt{n})}sim frac{1}{2n^{q+frac12}}$$



                        then refer to limit comparison test to conlcude that $q+frac12>1$.






                        share|cite|improve this answer


























                          1












                          1








                          1






                          From here



                          $$sum_{n=1}^{infty}{frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$$



                          we have that



                          $$frac{1}{n^q(sqrt{n+1}+sqrt{n})}sim frac{1}{2n^{q+frac12}}$$



                          then refer to limit comparison test to conlcude that $q+frac12>1$.






                          share|cite|improve this answer














                          From here



                          $$sum_{n=1}^{infty}{frac{1}{n^q(sqrt{n+1}+sqrt{n})}}$$



                          we have that



                          $$frac{1}{n^q(sqrt{n+1}+sqrt{n})}sim frac{1}{2n^{q+frac12}}$$



                          then refer to limit comparison test to conlcude that $q+frac12>1$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 29 '18 at 15:46

























                          answered Nov 29 '18 at 15:31









                          gimusigimusi

                          1




                          1






























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