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.






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    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






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      asked Dec 1 at 3:06









      suchanda adhikari

      516




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          2 Answers
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          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





















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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

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            active

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            active

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            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






























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