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







share|cite|improve this question



























    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
          3






          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





















          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





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018774%2fdetermine-all-q-in-mathbbq-so-that-sum-n-1-infty-frac-sqrtn1%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              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






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018774%2fdetermine-all-q-in-mathbbq-so-that-sum-n-1-infty-frac-sqrtn1%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              Le Mesnil-Réaume

                              Image
                              Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
                              Read more

                              Ida-Boy-Ed-Garten

                              Image
                              Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                              Read more

                              web3.py web3.isConnected() returns false always

                              Image
                              1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
                              Read more