Evaluate:$limlimits_{ntoinfty}sum_{k=0}^n frac1{k!}left(-frac12right)^k$












0












$begingroup$


Evaluate:$displaystylelim_{ntoinfty}sum_{k=0}^n frac1{k!}left(-frac12right)^k$



MY TRY:We know that $displaystylelim_{ntoinfty}sum_{k=0}^n frac1{k!}=e$ and
$displaystylelim_{ntoinfty}sum_{k=0}^n left(-frac12right)^k=frac 23$ but how can we evaluate the above$?$Thank you.



Note:The answer is $frac{1}{sqrt e}$






real-analysis sequences-and-series limits







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  • 3




    $begingroup$
    We have $e^x=sum_{k=0}^inftyfrac{x^k}{k!}$.
    $endgroup$
    – user296113
    Jan 14 '17 at 12:58










  • $begingroup$
    Possible duplicate of How does $e^0 = 1$ if you define $e^x = sum_{n = 0}^infty x^n/n!$
    $endgroup$
    – Guy Fsone
    Dec 29 '17 at 19:26










  • $begingroup$
    math.stackexchange.com/questions/441836/…
    $endgroup$
    – Guy Fsone
    Dec 29 '17 at 19:28
















0












$begingroup$


Evaluate:$displaystylelim_{ntoinfty}sum_{k=0}^n frac1{k!}left(-frac12right)^k$



MY TRY:We know that $displaystylelim_{ntoinfty}sum_{k=0}^n frac1{k!}=e$ and
$displaystylelim_{ntoinfty}sum_{k=0}^n left(-frac12right)^k=frac 23$ but how can we evaluate the above$?$Thank you.



Note:The answer is $frac{1}{sqrt e}$






real-analysis sequences-and-series limits







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    We have $e^x=sum_{k=0}^inftyfrac{x^k}{k!}$.
    $endgroup$
    – user296113
    Jan 14 '17 at 12:58










  • $begingroup$
    Possible duplicate of How does $e^0 = 1$ if you define $e^x = sum_{n = 0}^infty x^n/n!$
    $endgroup$
    – Guy Fsone
    Dec 29 '17 at 19:26










  • $begingroup$
    math.stackexchange.com/questions/441836/…
    $endgroup$
    – Guy Fsone
    Dec 29 '17 at 19:28














0












0








0





$begingroup$


Evaluate:$displaystylelim_{ntoinfty}sum_{k=0}^n frac1{k!}left(-frac12right)^k$



MY TRY:We know that $displaystylelim_{ntoinfty}sum_{k=0}^n frac1{k!}=e$ and
$displaystylelim_{ntoinfty}sum_{k=0}^n left(-frac12right)^k=frac 23$ but how can we evaluate the above$?$Thank you.



Note:The answer is $frac{1}{sqrt e}$






real-analysis sequences-and-series limits







share|cite|improve this question











$endgroup$




Evaluate:$displaystylelim_{ntoinfty}sum_{k=0}^n frac1{k!}left(-frac12right)^k$



MY TRY:We know that $displaystylelim_{ntoinfty}sum_{k=0}^n frac1{k!}=e$ and
$displaystylelim_{ntoinfty}sum_{k=0}^n left(-frac12right)^k=frac 23$ but how can we evaluate the above$?$Thank you.



Note:The answer is $frac{1}{sqrt e}$






real-analysis sequences-and-series limits




real-analysis sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 7:26









Martin Sleziak

44.8k10119273




44.8k10119273










asked Jan 14 '17 at 12:56









MatheMagicMatheMagic

1,4021617




1,4021617








  • 3




    $begingroup$
    We have $e^x=sum_{k=0}^inftyfrac{x^k}{k!}$.
    $endgroup$
    – user296113
    Jan 14 '17 at 12:58










  • $begingroup$
    Possible duplicate of How does $e^0 = 1$ if you define $e^x = sum_{n = 0}^infty x^n/n!$
    $endgroup$
    – Guy Fsone
    Dec 29 '17 at 19:26










  • $begingroup$
    math.stackexchange.com/questions/441836/…
    $endgroup$
    – Guy Fsone
    Dec 29 '17 at 19:28














  • 3




    $begingroup$
    We have $e^x=sum_{k=0}^inftyfrac{x^k}{k!}$.
    $endgroup$
    – user296113
    Jan 14 '17 at 12:58










  • $begingroup$
    Possible duplicate of How does $e^0 = 1$ if you define $e^x = sum_{n = 0}^infty x^n/n!$
    $endgroup$
    – Guy Fsone
    Dec 29 '17 at 19:26










  • $begingroup$
    math.stackexchange.com/questions/441836/…
    $endgroup$
    – Guy Fsone
    Dec 29 '17 at 19:28








3




3




$begingroup$
We have $e^x=sum_{k=0}^inftyfrac{x^k}{k!}$.
$endgroup$
– user296113
Jan 14 '17 at 12:58




$begingroup$
We have $e^x=sum_{k=0}^inftyfrac{x^k}{k!}$.
$endgroup$
– user296113
Jan 14 '17 at 12:58












$begingroup$
Possible duplicate of How does $e^0 = 1$ if you define $e^x = sum_{n = 0}^infty x^n/n!$
$endgroup$
– Guy Fsone
Dec 29 '17 at 19:26




$begingroup$
Possible duplicate of How does $e^0 = 1$ if you define $e^x = sum_{n = 0}^infty x^n/n!$
$endgroup$
– Guy Fsone
Dec 29 '17 at 19:26












$begingroup$
math.stackexchange.com/questions/441836/…
$endgroup$
– Guy Fsone
Dec 29 '17 at 19:28




$begingroup$
math.stackexchange.com/questions/441836/…
$endgroup$
– Guy Fsone
Dec 29 '17 at 19:28










2 Answers
2






active

oldest

votes


















3












$begingroup$

HINT:



$$e^x=sum_{r=0}^inftydfrac{x^r}{r!}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why the second hint?
    $endgroup$
    – Did
    Dec 7 '18 at 9:35










  • $begingroup$
    @Did, For $$displaystylelim_{ntoinfty}sum_{k=0}^n left(-frac12right)^k=frac 23$$ in the question
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 9:40






  • 1




    $begingroup$
    No, this is mentioned as a try and we know it leads nowhere. Adding it to your answer can only mislead the OP to think they need this fact to solve their question, while they do not.
    $endgroup$
    – Did
    Dec 7 '18 at 9:45










  • $begingroup$
    @Did, I added that formula so that OP can compare that with that of $e^x$
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 9:50



















1












$begingroup$

We have $displaystyle e^x=sum_{k=0}^inftyfrac{x^k}{k!}$.



So $displaystylelim_{ntoinfty}sum_{k=0}^n frac1{k!}left(-frac12right)^k=displaystylelim_{ntoinfty}sum_{k=0}^n frac{left(-frac12right)^k}{k!}=e^{frac {-1}{2}}=frac 1{sqrt e}$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    HINT:



    $$e^x=sum_{r=0}^inftydfrac{x^r}{r!}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Why the second hint?
      $endgroup$
      – Did
      Dec 7 '18 at 9:35










    • $begingroup$
      @Did, For $$displaystylelim_{ntoinfty}sum_{k=0}^n left(-frac12right)^k=frac 23$$ in the question
      $endgroup$
      – lab bhattacharjee
      Dec 7 '18 at 9:40






    • 1




      $begingroup$
      No, this is mentioned as a try and we know it leads nowhere. Adding it to your answer can only mislead the OP to think they need this fact to solve their question, while they do not.
      $endgroup$
      – Did
      Dec 7 '18 at 9:45










    • $begingroup$
      @Did, I added that formula so that OP can compare that with that of $e^x$
      $endgroup$
      – lab bhattacharjee
      Dec 7 '18 at 9:50
















    3












    $begingroup$

    HINT:



    $$e^x=sum_{r=0}^inftydfrac{x^r}{r!}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Why the second hint?
      $endgroup$
      – Did
      Dec 7 '18 at 9:35










    • $begingroup$
      @Did, For $$displaystylelim_{ntoinfty}sum_{k=0}^n left(-frac12right)^k=frac 23$$ in the question
      $endgroup$
      – lab bhattacharjee
      Dec 7 '18 at 9:40






    • 1




      $begingroup$
      No, this is mentioned as a try and we know it leads nowhere. Adding it to your answer can only mislead the OP to think they need this fact to solve their question, while they do not.
      $endgroup$
      – Did
      Dec 7 '18 at 9:45










    • $begingroup$
      @Did, I added that formula so that OP can compare that with that of $e^x$
      $endgroup$
      – lab bhattacharjee
      Dec 7 '18 at 9:50














    3












    3








    3





    $begingroup$

    HINT:



    $$e^x=sum_{r=0}^inftydfrac{x^r}{r!}$$






    share|cite|improve this answer











    $endgroup$



    HINT:



    $$e^x=sum_{r=0}^inftydfrac{x^r}{r!}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 7 '18 at 9:49

























    answered Jan 14 '17 at 12:58









    lab bhattacharjeelab bhattacharjee

    226k15158275




    226k15158275












    • $begingroup$
      Why the second hint?
      $endgroup$
      – Did
      Dec 7 '18 at 9:35










    • $begingroup$
      @Did, For $$displaystylelim_{ntoinfty}sum_{k=0}^n left(-frac12right)^k=frac 23$$ in the question
      $endgroup$
      – lab bhattacharjee
      Dec 7 '18 at 9:40






    • 1




      $begingroup$
      No, this is mentioned as a try and we know it leads nowhere. Adding it to your answer can only mislead the OP to think they need this fact to solve their question, while they do not.
      $endgroup$
      – Did
      Dec 7 '18 at 9:45










    • $begingroup$
      @Did, I added that formula so that OP can compare that with that of $e^x$
      $endgroup$
      – lab bhattacharjee
      Dec 7 '18 at 9:50


















    • $begingroup$
      Why the second hint?
      $endgroup$
      – Did
      Dec 7 '18 at 9:35










    • $begingroup$
      @Did, For $$displaystylelim_{ntoinfty}sum_{k=0}^n left(-frac12right)^k=frac 23$$ in the question
      $endgroup$
      – lab bhattacharjee
      Dec 7 '18 at 9:40






    • 1




      $begingroup$
      No, this is mentioned as a try and we know it leads nowhere. Adding it to your answer can only mislead the OP to think they need this fact to solve their question, while they do not.
      $endgroup$
      – Did
      Dec 7 '18 at 9:45










    • $begingroup$
      @Did, I added that formula so that OP can compare that with that of $e^x$
      $endgroup$
      – lab bhattacharjee
      Dec 7 '18 at 9:50
















    $begingroup$
    Why the second hint?
    $endgroup$
    – Did
    Dec 7 '18 at 9:35




    $begingroup$
    Why the second hint?
    $endgroup$
    – Did
    Dec 7 '18 at 9:35












    $begingroup$
    @Did, For $$displaystylelim_{ntoinfty}sum_{k=0}^n left(-frac12right)^k=frac 23$$ in the question
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 9:40




    $begingroup$
    @Did, For $$displaystylelim_{ntoinfty}sum_{k=0}^n left(-frac12right)^k=frac 23$$ in the question
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 9:40




    1




    1




    $begingroup$
    No, this is mentioned as a try and we know it leads nowhere. Adding it to your answer can only mislead the OP to think they need this fact to solve their question, while they do not.
    $endgroup$
    – Did
    Dec 7 '18 at 9:45




    $begingroup$
    No, this is mentioned as a try and we know it leads nowhere. Adding it to your answer can only mislead the OP to think they need this fact to solve their question, while they do not.
    $endgroup$
    – Did
    Dec 7 '18 at 9:45












    $begingroup$
    @Did, I added that formula so that OP can compare that with that of $e^x$
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 9:50




    $begingroup$
    @Did, I added that formula so that OP can compare that with that of $e^x$
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 9:50











    1












    $begingroup$

    We have $displaystyle e^x=sum_{k=0}^inftyfrac{x^k}{k!}$.



    So $displaystylelim_{ntoinfty}sum_{k=0}^n frac1{k!}left(-frac12right)^k=displaystylelim_{ntoinfty}sum_{k=0}^n frac{left(-frac12right)^k}{k!}=e^{frac {-1}{2}}=frac 1{sqrt e}$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      We have $displaystyle e^x=sum_{k=0}^inftyfrac{x^k}{k!}$.



      So $displaystylelim_{ntoinfty}sum_{k=0}^n frac1{k!}left(-frac12right)^k=displaystylelim_{ntoinfty}sum_{k=0}^n frac{left(-frac12right)^k}{k!}=e^{frac {-1}{2}}=frac 1{sqrt e}$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        We have $displaystyle e^x=sum_{k=0}^inftyfrac{x^k}{k!}$.



        So $displaystylelim_{ntoinfty}sum_{k=0}^n frac1{k!}left(-frac12right)^k=displaystylelim_{ntoinfty}sum_{k=0}^n frac{left(-frac12right)^k}{k!}=e^{frac {-1}{2}}=frac 1{sqrt e}$






        share|cite|improve this answer









        $endgroup$



        We have $displaystyle e^x=sum_{k=0}^inftyfrac{x^k}{k!}$.



        So $displaystylelim_{ntoinfty}sum_{k=0}^n frac1{k!}left(-frac12right)^k=displaystylelim_{ntoinfty}sum_{k=0}^n frac{left(-frac12right)^k}{k!}=e^{frac {-1}{2}}=frac 1{sqrt e}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 '17 at 13:17









        MatheMagicMatheMagic

        1,4021617




        1,4021617






























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