Alternating series example












-1












$begingroup$


I have a problem with rather simple example:



$$sumlimits_{n=0}^{infty} (-1)^{n+1}frac{10^n}{n!}$$



I have to tell if it's convergent. I know it is, but I don't know how to prove it. I was thinking of Leibniz's test, but this sequence doesn't decrease monotonically.



Thank you in advance.






sequences-and-series convergence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It is not necessary that the sequence ${a_n}$ of absolute values must decrease monotonically $forall n in mathbb{N}$.
    $endgroup$
    – Elman
    Dec 6 '18 at 22:32






  • 1




    $begingroup$
    The series converges absolutely since $sumlimits_{n=0}^{infty} left|(-1)^{n+1}frac{10^n}{n!}right|=sumlimits_{n=0}^{infty} frac{10^n}{n!}=e^{10}$.
    $endgroup$
    – Matt A Pelto
    Dec 6 '18 at 22:35












  • $begingroup$
    You just need that the sequence of absolute values eventually decreases monotonically to $0$. That is for sufficiently large $n$, $|a_{n+1}|le|a_n|$
    $endgroup$
    – saulspatz
    Dec 6 '18 at 22:44
















-1












$begingroup$


I have a problem with rather simple example:



$$sumlimits_{n=0}^{infty} (-1)^{n+1}frac{10^n}{n!}$$



I have to tell if it's convergent. I know it is, but I don't know how to prove it. I was thinking of Leibniz's test, but this sequence doesn't decrease monotonically.



Thank you in advance.






sequences-and-series convergence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It is not necessary that the sequence ${a_n}$ of absolute values must decrease monotonically $forall n in mathbb{N}$.
    $endgroup$
    – Elman
    Dec 6 '18 at 22:32






  • 1




    $begingroup$
    The series converges absolutely since $sumlimits_{n=0}^{infty} left|(-1)^{n+1}frac{10^n}{n!}right|=sumlimits_{n=0}^{infty} frac{10^n}{n!}=e^{10}$.
    $endgroup$
    – Matt A Pelto
    Dec 6 '18 at 22:35












  • $begingroup$
    You just need that the sequence of absolute values eventually decreases monotonically to $0$. That is for sufficiently large $n$, $|a_{n+1}|le|a_n|$
    $endgroup$
    – saulspatz
    Dec 6 '18 at 22:44














-1












-1








-1





$begingroup$


I have a problem with rather simple example:



$$sumlimits_{n=0}^{infty} (-1)^{n+1}frac{10^n}{n!}$$



I have to tell if it's convergent. I know it is, but I don't know how to prove it. I was thinking of Leibniz's test, but this sequence doesn't decrease monotonically.



Thank you in advance.






sequences-and-series convergence







share|cite|improve this question











$endgroup$




I have a problem with rather simple example:



$$sumlimits_{n=0}^{infty} (-1)^{n+1}frac{10^n}{n!}$$



I have to tell if it's convergent. I know it is, but I don't know how to prove it. I was thinking of Leibniz's test, but this sequence doesn't decrease monotonically.



Thank you in advance.






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 22:33









Masacroso

13.1k41747




13.1k41747










asked Dec 6 '18 at 22:26







user609637















  • 1




    $begingroup$
    It is not necessary that the sequence ${a_n}$ of absolute values must decrease monotonically $forall n in mathbb{N}$.
    $endgroup$
    – Elman
    Dec 6 '18 at 22:32






  • 1




    $begingroup$
    The series converges absolutely since $sumlimits_{n=0}^{infty} left|(-1)^{n+1}frac{10^n}{n!}right|=sumlimits_{n=0}^{infty} frac{10^n}{n!}=e^{10}$.
    $endgroup$
    – Matt A Pelto
    Dec 6 '18 at 22:35












  • $begingroup$
    You just need that the sequence of absolute values eventually decreases monotonically to $0$. That is for sufficiently large $n$, $|a_{n+1}|le|a_n|$
    $endgroup$
    – saulspatz
    Dec 6 '18 at 22:44














  • 1




    $begingroup$
    It is not necessary that the sequence ${a_n}$ of absolute values must decrease monotonically $forall n in mathbb{N}$.
    $endgroup$
    – Elman
    Dec 6 '18 at 22:32






  • 1




    $begingroup$
    The series converges absolutely since $sumlimits_{n=0}^{infty} left|(-1)^{n+1}frac{10^n}{n!}right|=sumlimits_{n=0}^{infty} frac{10^n}{n!}=e^{10}$.
    $endgroup$
    – Matt A Pelto
    Dec 6 '18 at 22:35












  • $begingroup$
    You just need that the sequence of absolute values eventually decreases monotonically to $0$. That is for sufficiently large $n$, $|a_{n+1}|le|a_n|$
    $endgroup$
    – saulspatz
    Dec 6 '18 at 22:44








1




1




$begingroup$
It is not necessary that the sequence ${a_n}$ of absolute values must decrease monotonically $forall n in mathbb{N}$.
$endgroup$
– Elman
Dec 6 '18 at 22:32




$begingroup$
It is not necessary that the sequence ${a_n}$ of absolute values must decrease monotonically $forall n in mathbb{N}$.
$endgroup$
– Elman
Dec 6 '18 at 22:32




1




1




$begingroup$
The series converges absolutely since $sumlimits_{n=0}^{infty} left|(-1)^{n+1}frac{10^n}{n!}right|=sumlimits_{n=0}^{infty} frac{10^n}{n!}=e^{10}$.
$endgroup$
– Matt A Pelto
Dec 6 '18 at 22:35






$begingroup$
The series converges absolutely since $sumlimits_{n=0}^{infty} left|(-1)^{n+1}frac{10^n}{n!}right|=sumlimits_{n=0}^{infty} frac{10^n}{n!}=e^{10}$.
$endgroup$
– Matt A Pelto
Dec 6 '18 at 22:35














$begingroup$
You just need that the sequence of absolute values eventually decreases monotonically to $0$. That is for sufficiently large $n$, $|a_{n+1}|le|a_n|$
$endgroup$
– saulspatz
Dec 6 '18 at 22:44




$begingroup$
You just need that the sequence of absolute values eventually decreases monotonically to $0$. That is for sufficiently large $n$, $|a_{n+1}|le|a_n|$
$endgroup$
– saulspatz
Dec 6 '18 at 22:44










2 Answers
2






active

oldest

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3












$begingroup$

We have that, by induction, for $n$ sufficiently large (base case $n=10$) and assuming (induction step)



$$frac{10^{n+1}}{(n+1)!}<frac{10^n}{n!}$$



we have



$$frac{10^{n+2}}{(n+2)!}=frac{10}{n+2}frac{10^{n+1}}{(n+1)!}<frac{10}{n+2}frac{10^n}{n!}<frac{10^{n+1}}{(n+1)!}$$



therefore $a_nto 0$ is strictly decreasing and we can refer to alternating series test.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    If we recall the identity
    $$e^x = sumlimits_{n=0}^{infty} frac{x^n}{n!} $$



    Then we can calculate (as the series above exists)
    $$-e^{-10} =-sumlimits_{n=0}^{infty}frac{(-10)^n}{n!} = sumlimits_{n=0}^{infty} (-1)^{n+1}frac{10^n}{n!} $$
    So it is a finite value and the series in question converges.






    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












      $begingroup$

      We have that, by induction, for $n$ sufficiently large (base case $n=10$) and assuming (induction step)



      $$frac{10^{n+1}}{(n+1)!}<frac{10^n}{n!}$$



      we have



      $$frac{10^{n+2}}{(n+2)!}=frac{10}{n+2}frac{10^{n+1}}{(n+1)!}<frac{10}{n+2}frac{10^n}{n!}<frac{10^{n+1}}{(n+1)!}$$



      therefore $a_nto 0$ is strictly decreasing and we can refer to alternating series test.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        We have that, by induction, for $n$ sufficiently large (base case $n=10$) and assuming (induction step)



        $$frac{10^{n+1}}{(n+1)!}<frac{10^n}{n!}$$



        we have



        $$frac{10^{n+2}}{(n+2)!}=frac{10}{n+2}frac{10^{n+1}}{(n+1)!}<frac{10}{n+2}frac{10^n}{n!}<frac{10^{n+1}}{(n+1)!}$$



        therefore $a_nto 0$ is strictly decreasing and we can refer to alternating series test.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          We have that, by induction, for $n$ sufficiently large (base case $n=10$) and assuming (induction step)



          $$frac{10^{n+1}}{(n+1)!}<frac{10^n}{n!}$$



          we have



          $$frac{10^{n+2}}{(n+2)!}=frac{10}{n+2}frac{10^{n+1}}{(n+1)!}<frac{10}{n+2}frac{10^n}{n!}<frac{10^{n+1}}{(n+1)!}$$



          therefore $a_nto 0$ is strictly decreasing and we can refer to alternating series test.






          share|cite|improve this answer









          $endgroup$



          We have that, by induction, for $n$ sufficiently large (base case $n=10$) and assuming (induction step)



          $$frac{10^{n+1}}{(n+1)!}<frac{10^n}{n!}$$



          we have



          $$frac{10^{n+2}}{(n+2)!}=frac{10}{n+2}frac{10^{n+1}}{(n+1)!}<frac{10}{n+2}frac{10^n}{n!}<frac{10^{n+1}}{(n+1)!}$$



          therefore $a_nto 0$ is strictly decreasing and we can refer to alternating series test.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 22:31









          gimusigimusi

          92.9k84494




          92.9k84494























              1












              $begingroup$

              If we recall the identity
              $$e^x = sumlimits_{n=0}^{infty} frac{x^n}{n!} $$



              Then we can calculate (as the series above exists)
              $$-e^{-10} =-sumlimits_{n=0}^{infty}frac{(-10)^n}{n!} = sumlimits_{n=0}^{infty} (-1)^{n+1}frac{10^n}{n!} $$
              So it is a finite value and the series in question converges.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                If we recall the identity
                $$e^x = sumlimits_{n=0}^{infty} frac{x^n}{n!} $$



                Then we can calculate (as the series above exists)
                $$-e^{-10} =-sumlimits_{n=0}^{infty}frac{(-10)^n}{n!} = sumlimits_{n=0}^{infty} (-1)^{n+1}frac{10^n}{n!} $$
                So it is a finite value and the series in question converges.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  If we recall the identity
                  $$e^x = sumlimits_{n=0}^{infty} frac{x^n}{n!} $$



                  Then we can calculate (as the series above exists)
                  $$-e^{-10} =-sumlimits_{n=0}^{infty}frac{(-10)^n}{n!} = sumlimits_{n=0}^{infty} (-1)^{n+1}frac{10^n}{n!} $$
                  So it is a finite value and the series in question converges.






                  share|cite|improve this answer









                  $endgroup$



                  If we recall the identity
                  $$e^x = sumlimits_{n=0}^{infty} frac{x^n}{n!} $$



                  Then we can calculate (as the series above exists)
                  $$-e^{-10} =-sumlimits_{n=0}^{infty}frac{(-10)^n}{n!} = sumlimits_{n=0}^{infty} (-1)^{n+1}frac{10^n}{n!} $$
                  So it is a finite value and the series in question converges.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 22:57









                  MaksimMaksim

                  65218




                  65218






























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