Discuss the convergence or divergence of $ sum_{n=1}^{infty} frac {n!} {e^{n^2}} $












0














I did the ratio test and showed that the limit of $ frac {a_{n+1}}{a_n} $ is zero as n tends to infinity So I concluded that the given series is convergent but the answer is divergent .
I checked again and again and I got the same limit zero . Any hint would be appreciated. Thanks in advance.






real-analysis







share|cite|improve this question



























    0














    I did the ratio test and showed that the limit of $ frac {a_{n+1}}{a_n} $ is zero as n tends to infinity So I concluded that the given series is convergent but the answer is divergent .
    I checked again and again and I got the same limit zero . Any hint would be appreciated. Thanks in advance.






    real-analysis







    share|cite|improve this question

























      0












      0








      0







      I did the ratio test and showed that the limit of $ frac {a_{n+1}}{a_n} $ is zero as n tends to infinity So I concluded that the given series is convergent but the answer is divergent .
      I checked again and again and I got the same limit zero . Any hint would be appreciated. Thanks in advance.






      real-analysis







      share|cite|improve this question













      I did the ratio test and showed that the limit of $ frac {a_{n+1}}{a_n} $ is zero as n tends to infinity So I concluded that the given series is convergent but the answer is divergent .
      I checked again and again and I got the same limit zero . Any hint would be appreciated. Thanks in advance.






      real-analysis




      real-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 1 at 3:06









      suchanda adhikari

      516




      516






















          2 Answers
          2






          active

          oldest

          votes


















          5














          Through the ratio test,



          $$frac{a_{n+1}}{a_n} = frac{(n+1)!}{e^{(n+1)^2}} frac{e^{n^2}}{n!}$$



          We note:



          $$(n+1)! = (n+1) cdot n!$$
          $$e^{(n+1)^2} = e^{n^2 + 2n + 1} = e^{n^2} cdot e^{2n} cdot e$$



          Thus,



          $$frac{(n+1)!}{e^{(n+1)^2}} frac{e^{n^2}}{n!} = frac{(n+1) cdot n!}{e^{n^2} cdot e^{2n} cdot e} frac{e^{n^2}}{n!}$$



          The $e^{n^2}$ and $n!$ terms cancel to give



          $$frac{(n+1) cdot n!}{e^{n^2} cdot e^{2n} cdot e} frac{e^{n^2}}{n!} = frac{n+1}{e^{2n} cdot e} = frac{n+1}{e^{2n+1}}$$



          Then, clearly, as $n to infty$, the ratio of successive terms approaches $0$.



          ... So I guess the question is, who or what is telling you the series doesn't converge? Looks like it does, and Wolfram Alpha (https://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+infinity+of+n!+%2F+e%5E(n%5E2)) also confirms as much, though Wolfram Alpha can be iffy with some series so it's not foolproof.



          Granted it's been a while since I dealt with series through the ratio test so I could easily be wrong.






          share|cite|improve this answer





















          • Actually it is an exercise from introduction to real analysis by Bartle and Sherbert. .and first I thought the answer was wrong but then I checked the latest edition of the book ..and the answer was still divergent. .so I thought I am wrong and there is some kind of trick I am unable to find and am extremely sorry ..thanks for answering.
            – suchanda adhikari
            Dec 1 at 3:38






          • 4




            I would venture then that the book just plain has an error! It happens - and they even persist over multiple generations of edition. Factorial is asymptotic to exp(n.log(n)), so exp(n^2) has a very much greater order of growth. There is no doubt that the ratio of the former to the latter tends to zero very quickly.
            – AmbretteOrrisey
            Dec 1 at 4:12





















          1














          Alternatevly You can do more efficently using root test and using Stirling Approximation



          $n!=sqrt(2pi e)(n/e)^n$



          So apply n th root test



          then



          $frac{(sqrt(2pi e))^{1/n}(n/e)}{e^n}=n/e^{n+1}<1 forall n$



          SO converges






          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%2f3020936%2fdiscuss-the-convergence-or-divergence-of-sum-n-1-infty-frac-n-en%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5














            Through the ratio test,



            $$frac{a_{n+1}}{a_n} = frac{(n+1)!}{e^{(n+1)^2}} frac{e^{n^2}}{n!}$$



            We note:



            $$(n+1)! = (n+1) cdot n!$$
            $$e^{(n+1)^2} = e^{n^2 + 2n + 1} = e^{n^2} cdot e^{2n} cdot e$$



            Thus,



            $$frac{(n+1)!}{e^{(n+1)^2}} frac{e^{n^2}}{n!} = frac{(n+1) cdot n!}{e^{n^2} cdot e^{2n} cdot e} frac{e^{n^2}}{n!}$$



            The $e^{n^2}$ and $n!$ terms cancel to give



            $$frac{(n+1) cdot n!}{e^{n^2} cdot e^{2n} cdot e} frac{e^{n^2}}{n!} = frac{n+1}{e^{2n} cdot e} = frac{n+1}{e^{2n+1}}$$



            Then, clearly, as $n to infty$, the ratio of successive terms approaches $0$.



            ... So I guess the question is, who or what is telling you the series doesn't converge? Looks like it does, and Wolfram Alpha (https://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+infinity+of+n!+%2F+e%5E(n%5E2)) also confirms as much, though Wolfram Alpha can be iffy with some series so it's not foolproof.



            Granted it's been a while since I dealt with series through the ratio test so I could easily be wrong.






            share|cite|improve this answer





















            • Actually it is an exercise from introduction to real analysis by Bartle and Sherbert. .and first I thought the answer was wrong but then I checked the latest edition of the book ..and the answer was still divergent. .so I thought I am wrong and there is some kind of trick I am unable to find and am extremely sorry ..thanks for answering.
              – suchanda adhikari
              Dec 1 at 3:38






            • 4




              I would venture then that the book just plain has an error! It happens - and they even persist over multiple generations of edition. Factorial is asymptotic to exp(n.log(n)), so exp(n^2) has a very much greater order of growth. There is no doubt that the ratio of the former to the latter tends to zero very quickly.
              – AmbretteOrrisey
              Dec 1 at 4:12


















            5














            Through the ratio test,



            $$frac{a_{n+1}}{a_n} = frac{(n+1)!}{e^{(n+1)^2}} frac{e^{n^2}}{n!}$$



            We note:



            $$(n+1)! = (n+1) cdot n!$$
            $$e^{(n+1)^2} = e^{n^2 + 2n + 1} = e^{n^2} cdot e^{2n} cdot e$$



            Thus,



            $$frac{(n+1)!}{e^{(n+1)^2}} frac{e^{n^2}}{n!} = frac{(n+1) cdot n!}{e^{n^2} cdot e^{2n} cdot e} frac{e^{n^2}}{n!}$$



            The $e^{n^2}$ and $n!$ terms cancel to give



            $$frac{(n+1) cdot n!}{e^{n^2} cdot e^{2n} cdot e} frac{e^{n^2}}{n!} = frac{n+1}{e^{2n} cdot e} = frac{n+1}{e^{2n+1}}$$



            Then, clearly, as $n to infty$, the ratio of successive terms approaches $0$.



            ... So I guess the question is, who or what is telling you the series doesn't converge? Looks like it does, and Wolfram Alpha (https://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+infinity+of+n!+%2F+e%5E(n%5E2)) also confirms as much, though Wolfram Alpha can be iffy with some series so it's not foolproof.



            Granted it's been a while since I dealt with series through the ratio test so I could easily be wrong.






            share|cite|improve this answer





















            • Actually it is an exercise from introduction to real analysis by Bartle and Sherbert. .and first I thought the answer was wrong but then I checked the latest edition of the book ..and the answer was still divergent. .so I thought I am wrong and there is some kind of trick I am unable to find and am extremely sorry ..thanks for answering.
              – suchanda adhikari
              Dec 1 at 3:38






            • 4




              I would venture then that the book just plain has an error! It happens - and they even persist over multiple generations of edition. Factorial is asymptotic to exp(n.log(n)), so exp(n^2) has a very much greater order of growth. There is no doubt that the ratio of the former to the latter tends to zero very quickly.
              – AmbretteOrrisey
              Dec 1 at 4:12
















            5












            5








            5






            Through the ratio test,



            $$frac{a_{n+1}}{a_n} = frac{(n+1)!}{e^{(n+1)^2}} frac{e^{n^2}}{n!}$$



            We note:



            $$(n+1)! = (n+1) cdot n!$$
            $$e^{(n+1)^2} = e^{n^2 + 2n + 1} = e^{n^2} cdot e^{2n} cdot e$$



            Thus,



            $$frac{(n+1)!}{e^{(n+1)^2}} frac{e^{n^2}}{n!} = frac{(n+1) cdot n!}{e^{n^2} cdot e^{2n} cdot e} frac{e^{n^2}}{n!}$$



            The $e^{n^2}$ and $n!$ terms cancel to give



            $$frac{(n+1) cdot n!}{e^{n^2} cdot e^{2n} cdot e} frac{e^{n^2}}{n!} = frac{n+1}{e^{2n} cdot e} = frac{n+1}{e^{2n+1}}$$



            Then, clearly, as $n to infty$, the ratio of successive terms approaches $0$.



            ... So I guess the question is, who or what is telling you the series doesn't converge? Looks like it does, and Wolfram Alpha (https://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+infinity+of+n!+%2F+e%5E(n%5E2)) also confirms as much, though Wolfram Alpha can be iffy with some series so it's not foolproof.



            Granted it's been a while since I dealt with series through the ratio test so I could easily be wrong.






            share|cite|improve this answer












            Through the ratio test,



            $$frac{a_{n+1}}{a_n} = frac{(n+1)!}{e^{(n+1)^2}} frac{e^{n^2}}{n!}$$



            We note:



            $$(n+1)! = (n+1) cdot n!$$
            $$e^{(n+1)^2} = e^{n^2 + 2n + 1} = e^{n^2} cdot e^{2n} cdot e$$



            Thus,



            $$frac{(n+1)!}{e^{(n+1)^2}} frac{e^{n^2}}{n!} = frac{(n+1) cdot n!}{e^{n^2} cdot e^{2n} cdot e} frac{e^{n^2}}{n!}$$



            The $e^{n^2}$ and $n!$ terms cancel to give



            $$frac{(n+1) cdot n!}{e^{n^2} cdot e^{2n} cdot e} frac{e^{n^2}}{n!} = frac{n+1}{e^{2n} cdot e} = frac{n+1}{e^{2n+1}}$$



            Then, clearly, as $n to infty$, the ratio of successive terms approaches $0$.



            ... So I guess the question is, who or what is telling you the series doesn't converge? Looks like it does, and Wolfram Alpha (https://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+infinity+of+n!+%2F+e%5E(n%5E2)) also confirms as much, though Wolfram Alpha can be iffy with some series so it's not foolproof.



            Granted it's been a while since I dealt with series through the ratio test so I could easily be wrong.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 1 at 3:17









            Eevee Trainer

            4,389630




            4,389630












            • Actually it is an exercise from introduction to real analysis by Bartle and Sherbert. .and first I thought the answer was wrong but then I checked the latest edition of the book ..and the answer was still divergent. .so I thought I am wrong and there is some kind of trick I am unable to find and am extremely sorry ..thanks for answering.
              – suchanda adhikari
              Dec 1 at 3:38






            • 4




              I would venture then that the book just plain has an error! It happens - and they even persist over multiple generations of edition. Factorial is asymptotic to exp(n.log(n)), so exp(n^2) has a very much greater order of growth. There is no doubt that the ratio of the former to the latter tends to zero very quickly.
              – AmbretteOrrisey
              Dec 1 at 4:12




















            • Actually it is an exercise from introduction to real analysis by Bartle and Sherbert. .and first I thought the answer was wrong but then I checked the latest edition of the book ..and the answer was still divergent. .so I thought I am wrong and there is some kind of trick I am unable to find and am extremely sorry ..thanks for answering.
              – suchanda adhikari
              Dec 1 at 3:38






            • 4




              I would venture then that the book just plain has an error! It happens - and they even persist over multiple generations of edition. Factorial is asymptotic to exp(n.log(n)), so exp(n^2) has a very much greater order of growth. There is no doubt that the ratio of the former to the latter tends to zero very quickly.
              – AmbretteOrrisey
              Dec 1 at 4:12


















            Actually it is an exercise from introduction to real analysis by Bartle and Sherbert. .and first I thought the answer was wrong but then I checked the latest edition of the book ..and the answer was still divergent. .so I thought I am wrong and there is some kind of trick I am unable to find and am extremely sorry ..thanks for answering.
            – suchanda adhikari
            Dec 1 at 3:38




            Actually it is an exercise from introduction to real analysis by Bartle and Sherbert. .and first I thought the answer was wrong but then I checked the latest edition of the book ..and the answer was still divergent. .so I thought I am wrong and there is some kind of trick I am unable to find and am extremely sorry ..thanks for answering.
            – suchanda adhikari
            Dec 1 at 3:38




            4




            4




            I would venture then that the book just plain has an error! It happens - and they even persist over multiple generations of edition. Factorial is asymptotic to exp(n.log(n)), so exp(n^2) has a very much greater order of growth. There is no doubt that the ratio of the former to the latter tends to zero very quickly.
            – AmbretteOrrisey
            Dec 1 at 4:12






            I would venture then that the book just plain has an error! It happens - and they even persist over multiple generations of edition. Factorial is asymptotic to exp(n.log(n)), so exp(n^2) has a very much greater order of growth. There is no doubt that the ratio of the former to the latter tends to zero very quickly.
            – AmbretteOrrisey
            Dec 1 at 4:12













            1














            Alternatevly You can do more efficently using root test and using Stirling Approximation



            $n!=sqrt(2pi e)(n/e)^n$



            So apply n th root test



            then



            $frac{(sqrt(2pi e))^{1/n}(n/e)}{e^n}=n/e^{n+1}<1 forall n$



            SO converges






            share|cite|improve this answer


























              1














              Alternatevly You can do more efficently using root test and using Stirling Approximation



              $n!=sqrt(2pi e)(n/e)^n$



              So apply n th root test



              then



              $frac{(sqrt(2pi e))^{1/n}(n/e)}{e^n}=n/e^{n+1}<1 forall n$



              SO converges






              share|cite|improve this answer
























                1












                1








                1






                Alternatevly You can do more efficently using root test and using Stirling Approximation



                $n!=sqrt(2pi e)(n/e)^n$



                So apply n th root test



                then



                $frac{(sqrt(2pi e))^{1/n}(n/e)}{e^n}=n/e^{n+1}<1 forall n$



                SO converges






                share|cite|improve this answer












                Alternatevly You can do more efficently using root test and using Stirling Approximation



                $n!=sqrt(2pi e)(n/e)^n$



                So apply n th root test



                then



                $frac{(sqrt(2pi e))^{1/n}(n/e)}{e^n}=n/e^{n+1}<1 forall n$



                SO converges







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 1 at 4:32









                Shubham

                1,5851519




                1,5851519






























                    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%2f3020936%2fdiscuss-the-convergence-or-divergence-of-sum-n-1-infty-frac-n-en%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

                    To store a contact into the json file from server.js file using a class in NodeJS

                    Image
                    0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
                    Read more

                    Redirect URL with Chrome Remote Debugging Android Devices

                    Image
                    1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
                    Read more

                    Dieringhausen

                    Image
                    Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
                    Read more