partial n-th sum of the infinity sequence and converging












-1












$begingroup$


I am suppose to find partial n-th sum of the infinity sequence and determine if it converge.



My sequence is $$ sum_{ n=1}^{infty} ln left(1+frac{1}{n}right) $$



My solution is :



I find out that first terms are:$ ln (2), lnfrac{3}{2}, ln frac{4}{3} $



Then I wrote that: $ s_{n}=(ln(2)+ln(frac{3}{2})+lnfrac{4}{3}+...ln(1+frac{1}{n}) ) $



I know that, it is not enough, but I have no idea, what to do next, or if this is true.



Thanks a lot for any advice.










share|cite|improve this question











$endgroup$

















    -1












    $begingroup$


    I am suppose to find partial n-th sum of the infinity sequence and determine if it converge.



    My sequence is $$ sum_{ n=1}^{infty} ln left(1+frac{1}{n}right) $$



    My solution is :



    I find out that first terms are:$ ln (2), lnfrac{3}{2}, ln frac{4}{3} $



    Then I wrote that: $ s_{n}=(ln(2)+ln(frac{3}{2})+lnfrac{4}{3}+...ln(1+frac{1}{n}) ) $



    I know that, it is not enough, but I have no idea, what to do next, or if this is true.



    Thanks a lot for any advice.










    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$


      I am suppose to find partial n-th sum of the infinity sequence and determine if it converge.



      My sequence is $$ sum_{ n=1}^{infty} ln left(1+frac{1}{n}right) $$



      My solution is :



      I find out that first terms are:$ ln (2), lnfrac{3}{2}, ln frac{4}{3} $



      Then I wrote that: $ s_{n}=(ln(2)+ln(frac{3}{2})+lnfrac{4}{3}+...ln(1+frac{1}{n}) ) $



      I know that, it is not enough, but I have no idea, what to do next, or if this is true.



      Thanks a lot for any advice.










      share|cite|improve this question











      $endgroup$




      I am suppose to find partial n-th sum of the infinity sequence and determine if it converge.



      My sequence is $$ sum_{ n=1}^{infty} ln left(1+frac{1}{n}right) $$



      My solution is :



      I find out that first terms are:$ ln (2), lnfrac{3}{2}, ln frac{4}{3} $



      Then I wrote that: $ s_{n}=(ln(2)+ln(frac{3}{2})+lnfrac{4}{3}+...ln(1+frac{1}{n}) ) $



      I know that, it is not enough, but I have no idea, what to do next, or if this is true.



      Thanks a lot for any advice.







      real-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 2 '18 at 17:10









      Larry

      2,27231028




      2,27231028










      asked Dec 2 '18 at 16:39









      ShelleyShelley

      92




      92






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          $$sum_{k=1}^n ln(1+1/k)=sum_{k=1}^nlnleft(frac{k+1}{k}right) =lnprod_{k=1}^n frac{k+1}{k} =lnfrac{(n+1)!}{n!}=ln(n+1)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I do not understand, why did you use factorial?
            $endgroup$
            – Shelley
            Dec 2 '18 at 16:53










          • $begingroup$
            Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
            $endgroup$
            – Tito Eliatron
            Dec 2 '18 at 16:56



















          0












          $begingroup$

          $$lnleft(frac{k+1}{k}right)=ln(k+1)-ln(k)$$
          $$sum_{k=1}^{n} ln left(1+frac{1}{k}right) = sum_{k=1}^{n}lnleft(frac{k+1}{k}right)=sum_{k=1}^{n}ln(k+1)-ln(k)=ln(n+1)$$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            As $xto 0,$



            $$frac{ln(1+x)}{x}= frac{ln(1+x)-ln 1}{x-0}to ln'(1)=1.$$



            This shows $[ln(1+1/n)]/(1/n)to 1.$ We know $sum 1/n$ diverges, so by the limit comparison test, $sum ln(1+1/n)$ diverges.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
              $endgroup$
              – zhw.
              Dec 2 '18 at 17:18













            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%2f3022860%2fpartial-n-th-sum-of-the-infinity-sequence-and-converging%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









            1












            $begingroup$

            $$sum_{k=1}^n ln(1+1/k)=sum_{k=1}^nlnleft(frac{k+1}{k}right) =lnprod_{k=1}^n frac{k+1}{k} =lnfrac{(n+1)!}{n!}=ln(n+1)$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I do not understand, why did you use factorial?
              $endgroup$
              – Shelley
              Dec 2 '18 at 16:53










            • $begingroup$
              Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
              $endgroup$
              – Tito Eliatron
              Dec 2 '18 at 16:56
















            1












            $begingroup$

            $$sum_{k=1}^n ln(1+1/k)=sum_{k=1}^nlnleft(frac{k+1}{k}right) =lnprod_{k=1}^n frac{k+1}{k} =lnfrac{(n+1)!}{n!}=ln(n+1)$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I do not understand, why did you use factorial?
              $endgroup$
              – Shelley
              Dec 2 '18 at 16:53










            • $begingroup$
              Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
              $endgroup$
              – Tito Eliatron
              Dec 2 '18 at 16:56














            1












            1








            1





            $begingroup$

            $$sum_{k=1}^n ln(1+1/k)=sum_{k=1}^nlnleft(frac{k+1}{k}right) =lnprod_{k=1}^n frac{k+1}{k} =lnfrac{(n+1)!}{n!}=ln(n+1)$$






            share|cite|improve this answer









            $endgroup$



            $$sum_{k=1}^n ln(1+1/k)=sum_{k=1}^nlnleft(frac{k+1}{k}right) =lnprod_{k=1}^n frac{k+1}{k} =lnfrac{(n+1)!}{n!}=ln(n+1)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 2 '18 at 16:44









            Tito EliatronTito Eliatron

            1,446622




            1,446622












            • $begingroup$
              I do not understand, why did you use factorial?
              $endgroup$
              – Shelley
              Dec 2 '18 at 16:53










            • $begingroup$
              Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
              $endgroup$
              – Tito Eliatron
              Dec 2 '18 at 16:56


















            • $begingroup$
              I do not understand, why did you use factorial?
              $endgroup$
              – Shelley
              Dec 2 '18 at 16:53










            • $begingroup$
              Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
              $endgroup$
              – Tito Eliatron
              Dec 2 '18 at 16:56
















            $begingroup$
            I do not understand, why did you use factorial?
            $endgroup$
            – Shelley
            Dec 2 '18 at 16:53




            $begingroup$
            I do not understand, why did you use factorial?
            $endgroup$
            – Shelley
            Dec 2 '18 at 16:53












            $begingroup$
            Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
            $endgroup$
            – Tito Eliatron
            Dec 2 '18 at 16:56




            $begingroup$
            Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
            $endgroup$
            – Tito Eliatron
            Dec 2 '18 at 16:56











            0












            $begingroup$

            $$lnleft(frac{k+1}{k}right)=ln(k+1)-ln(k)$$
            $$sum_{k=1}^{n} ln left(1+frac{1}{k}right) = sum_{k=1}^{n}lnleft(frac{k+1}{k}right)=sum_{k=1}^{n}ln(k+1)-ln(k)=ln(n+1)$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              $$lnleft(frac{k+1}{k}right)=ln(k+1)-ln(k)$$
              $$sum_{k=1}^{n} ln left(1+frac{1}{k}right) = sum_{k=1}^{n}lnleft(frac{k+1}{k}right)=sum_{k=1}^{n}ln(k+1)-ln(k)=ln(n+1)$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                $$lnleft(frac{k+1}{k}right)=ln(k+1)-ln(k)$$
                $$sum_{k=1}^{n} ln left(1+frac{1}{k}right) = sum_{k=1}^{n}lnleft(frac{k+1}{k}right)=sum_{k=1}^{n}ln(k+1)-ln(k)=ln(n+1)$$






                share|cite|improve this answer









                $endgroup$



                $$lnleft(frac{k+1}{k}right)=ln(k+1)-ln(k)$$
                $$sum_{k=1}^{n} ln left(1+frac{1}{k}right) = sum_{k=1}^{n}lnleft(frac{k+1}{k}right)=sum_{k=1}^{n}ln(k+1)-ln(k)=ln(n+1)$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 '18 at 16:57









                LarryLarry

                2,27231028




                2,27231028























                    0












                    $begingroup$

                    As $xto 0,$



                    $$frac{ln(1+x)}{x}= frac{ln(1+x)-ln 1}{x-0}to ln'(1)=1.$$



                    This shows $[ln(1+1/n)]/(1/n)to 1.$ We know $sum 1/n$ diverges, so by the limit comparison test, $sum ln(1+1/n)$ diverges.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
                      $endgroup$
                      – zhw.
                      Dec 2 '18 at 17:18


















                    0












                    $begingroup$

                    As $xto 0,$



                    $$frac{ln(1+x)}{x}= frac{ln(1+x)-ln 1}{x-0}to ln'(1)=1.$$



                    This shows $[ln(1+1/n)]/(1/n)to 1.$ We know $sum 1/n$ diverges, so by the limit comparison test, $sum ln(1+1/n)$ diverges.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
                      $endgroup$
                      – zhw.
                      Dec 2 '18 at 17:18
















                    0












                    0








                    0





                    $begingroup$

                    As $xto 0,$



                    $$frac{ln(1+x)}{x}= frac{ln(1+x)-ln 1}{x-0}to ln'(1)=1.$$



                    This shows $[ln(1+1/n)]/(1/n)to 1.$ We know $sum 1/n$ diverges, so by the limit comparison test, $sum ln(1+1/n)$ diverges.






                    share|cite|improve this answer









                    $endgroup$



                    As $xto 0,$



                    $$frac{ln(1+x)}{x}= frac{ln(1+x)-ln 1}{x-0}to ln'(1)=1.$$



                    This shows $[ln(1+1/n)]/(1/n)to 1.$ We know $sum 1/n$ diverges, so by the limit comparison test, $sum ln(1+1/n)$ diverges.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 2 '18 at 17:15









                    zhw.zhw.

                    72.2k43175




                    72.2k43175












                    • $begingroup$
                      More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
                      $endgroup$
                      – zhw.
                      Dec 2 '18 at 17:18




















                    • $begingroup$
                      More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
                      $endgroup$
                      – zhw.
                      Dec 2 '18 at 17:18


















                    $begingroup$
                    More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
                    $endgroup$
                    – zhw.
                    Dec 2 '18 at 17:18






                    $begingroup$
                    More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
                    $endgroup$
                    – zhw.
                    Dec 2 '18 at 17:18




















                    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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022860%2fpartial-n-th-sum-of-the-infinity-sequence-and-converging%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

                    Plaza Victoria

                    In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

                    How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...