Show that $S_n=1+{xover1!}+{x^2over2!}+cdots+{x^nover n!}$ converges for $ninBbb N, x inBbb R$ without using...












6












$begingroup$



Given a sequence ${S_n}$, $ninBbb N$:
$$
S_n=1+{xover1!}+{x^2over2!}+cdots+{x^nover n!}
$$
Prove that $S_n$ converges for all $xinBbb R$.




Please note that i know $S_n$ is a simple Taylor series for $e^x$, the case is I'm not supposed to know (and/or use) that fact when solving this problem since derivatives have not been defined yet.



I've started with a simpler case assuming $x = 1$. So the sequence becomes:
$$
S_n = 1+{1over1!}+{1over2!}+cdots+{1over n!}
$$



It seems reasonable to use Cauchy Criterion for the sequence. Namely suppose $m>n$, then we want to show:
$$
|x_n - x_m| < epsilon \
|x_m - x_n| = left|sum_{k=n+1}^m {1over k!}right|
$$
Lets try to synthetically bound the sum by:
$$
begin{align}
{1over k!} &= {1over k!}left(1 - {1over k} + {1over k}right) \
&le {1over k!}left(1 + {1over k-1} - {1over k}right) \
&= {1over k!}left({kover k -1} - {1over k}right) \
&= {1over (k - 1)(k-1)!} - {1over kcdot k!}
end{align}
$$



By telescoping we obtain:
$$
left|sum_{k=n+1}^m {1over k!}right| le left|{1over ncdot n!} - {1over mcdot m!}right|
$$



Given $m>n$ and $n,m in Bbb N$:
$$
left|{1over ncdot n!} - {1over mcdot m!}right| le left|1over ncdot n!right| = {1 over ncdot n!}
$$



Applying the limit to $|x_m - x_n|$ one may obtain:
$$
0 le lim_{ntoinfty}|x_{n+p} - x_n| le lim_{ntoinfty} {1over ncdot n!} = 0
$$



So squeezing $|x_m - x_n|$ gives:
$$
lim_{ntoinfty}|x_{n+p} - x_n| = 0, pin Bbb N
$$



Which would eventually mean:
$$
|x_m - x_n| < epsilon
$$



However I'm not sure how to find the index $N_epsilon$ from which the inequality becomes true since the expression for the upper bound involves a factorial.



If we now put $x = x_0 in Bbb R$:
$$
0 le lim_{ntoinfty}|x_{n+p} - x_n| le lim_{ntoinfty} {x_0over ncdot n!} = 0
$$
Which doesn't influence the value of the limit.




There are three questions in my mind:




  1. Is the overall reasoning valid?

  2. Is it possible to find a closed form of $N(epsilon)$, such that $n, m > N_epsilon implies |x_n - x_m| < epsilon$?

  3. Should I consider two cases for $xge 0$ and $x<0$




Thank you!






calculus sequences-and-series limits proof-verification cauchy-sequences







share|cite|improve this question











$endgroup$

















    6












    $begingroup$



    Given a sequence ${S_n}$, $ninBbb N$:
    $$
    S_n=1+{xover1!}+{x^2over2!}+cdots+{x^nover n!}
    $$
    Prove that $S_n$ converges for all $xinBbb R$.




    Please note that i know $S_n$ is a simple Taylor series for $e^x$, the case is I'm not supposed to know (and/or use) that fact when solving this problem since derivatives have not been defined yet.



    I've started with a simpler case assuming $x = 1$. So the sequence becomes:
    $$
    S_n = 1+{1over1!}+{1over2!}+cdots+{1over n!}
    $$



    It seems reasonable to use Cauchy Criterion for the sequence. Namely suppose $m>n$, then we want to show:
    $$
    |x_n - x_m| < epsilon \
    |x_m - x_n| = left|sum_{k=n+1}^m {1over k!}right|
    $$
    Lets try to synthetically bound the sum by:
    $$
    begin{align}
    {1over k!} &= {1over k!}left(1 - {1over k} + {1over k}right) \
    &le {1over k!}left(1 + {1over k-1} - {1over k}right) \
    &= {1over k!}left({kover k -1} - {1over k}right) \
    &= {1over (k - 1)(k-1)!} - {1over kcdot k!}
    end{align}
    $$



    By telescoping we obtain:
    $$
    left|sum_{k=n+1}^m {1over k!}right| le left|{1over ncdot n!} - {1over mcdot m!}right|
    $$



    Given $m>n$ and $n,m in Bbb N$:
    $$
    left|{1over ncdot n!} - {1over mcdot m!}right| le left|1over ncdot n!right| = {1 over ncdot n!}
    $$



    Applying the limit to $|x_m - x_n|$ one may obtain:
    $$
    0 le lim_{ntoinfty}|x_{n+p} - x_n| le lim_{ntoinfty} {1over ncdot n!} = 0
    $$



    So squeezing $|x_m - x_n|$ gives:
    $$
    lim_{ntoinfty}|x_{n+p} - x_n| = 0, pin Bbb N
    $$



    Which would eventually mean:
    $$
    |x_m - x_n| < epsilon
    $$



    However I'm not sure how to find the index $N_epsilon$ from which the inequality becomes true since the expression for the upper bound involves a factorial.



    If we now put $x = x_0 in Bbb R$:
    $$
    0 le lim_{ntoinfty}|x_{n+p} - x_n| le lim_{ntoinfty} {x_0over ncdot n!} = 0
    $$
    Which doesn't influence the value of the limit.




    There are three questions in my mind:




    1. Is the overall reasoning valid?

    2. Is it possible to find a closed form of $N(epsilon)$, such that $n, m > N_epsilon implies |x_n - x_m| < epsilon$?

    3. Should I consider two cases for $xge 0$ and $x<0$




    Thank you!






    calculus sequences-and-series limits proof-verification cauchy-sequences







    share|cite|improve this question











    $endgroup$















      6












      6








      6





      $begingroup$



      Given a sequence ${S_n}$, $ninBbb N$:
      $$
      S_n=1+{xover1!}+{x^2over2!}+cdots+{x^nover n!}
      $$
      Prove that $S_n$ converges for all $xinBbb R$.




      Please note that i know $S_n$ is a simple Taylor series for $e^x$, the case is I'm not supposed to know (and/or use) that fact when solving this problem since derivatives have not been defined yet.



      I've started with a simpler case assuming $x = 1$. So the sequence becomes:
      $$
      S_n = 1+{1over1!}+{1over2!}+cdots+{1over n!}
      $$



      It seems reasonable to use Cauchy Criterion for the sequence. Namely suppose $m>n$, then we want to show:
      $$
      |x_n - x_m| < epsilon \
      |x_m - x_n| = left|sum_{k=n+1}^m {1over k!}right|
      $$
      Lets try to synthetically bound the sum by:
      $$
      begin{align}
      {1over k!} &= {1over k!}left(1 - {1over k} + {1over k}right) \
      &le {1over k!}left(1 + {1over k-1} - {1over k}right) \
      &= {1over k!}left({kover k -1} - {1over k}right) \
      &= {1over (k - 1)(k-1)!} - {1over kcdot k!}
      end{align}
      $$



      By telescoping we obtain:
      $$
      left|sum_{k=n+1}^m {1over k!}right| le left|{1over ncdot n!} - {1over mcdot m!}right|
      $$



      Given $m>n$ and $n,m in Bbb N$:
      $$
      left|{1over ncdot n!} - {1over mcdot m!}right| le left|1over ncdot n!right| = {1 over ncdot n!}
      $$



      Applying the limit to $|x_m - x_n|$ one may obtain:
      $$
      0 le lim_{ntoinfty}|x_{n+p} - x_n| le lim_{ntoinfty} {1over ncdot n!} = 0
      $$



      So squeezing $|x_m - x_n|$ gives:
      $$
      lim_{ntoinfty}|x_{n+p} - x_n| = 0, pin Bbb N
      $$



      Which would eventually mean:
      $$
      |x_m - x_n| < epsilon
      $$



      However I'm not sure how to find the index $N_epsilon$ from which the inequality becomes true since the expression for the upper bound involves a factorial.



      If we now put $x = x_0 in Bbb R$:
      $$
      0 le lim_{ntoinfty}|x_{n+p} - x_n| le lim_{ntoinfty} {x_0over ncdot n!} = 0
      $$
      Which doesn't influence the value of the limit.




      There are three questions in my mind:




      1. Is the overall reasoning valid?

      2. Is it possible to find a closed form of $N(epsilon)$, such that $n, m > N_epsilon implies |x_n - x_m| < epsilon$?

      3. Should I consider two cases for $xge 0$ and $x<0$




      Thank you!






      calculus sequences-and-series limits proof-verification cauchy-sequences







      share|cite|improve this question











      $endgroup$





      Given a sequence ${S_n}$, $ninBbb N$:
      $$
      S_n=1+{xover1!}+{x^2over2!}+cdots+{x^nover n!}
      $$
      Prove that $S_n$ converges for all $xinBbb R$.




      Please note that i know $S_n$ is a simple Taylor series for $e^x$, the case is I'm not supposed to know (and/or use) that fact when solving this problem since derivatives have not been defined yet.



      I've started with a simpler case assuming $x = 1$. So the sequence becomes:
      $$
      S_n = 1+{1over1!}+{1over2!}+cdots+{1over n!}
      $$



      It seems reasonable to use Cauchy Criterion for the sequence. Namely suppose $m>n$, then we want to show:
      $$
      |x_n - x_m| < epsilon \
      |x_m - x_n| = left|sum_{k=n+1}^m {1over k!}right|
      $$
      Lets try to synthetically bound the sum by:
      $$
      begin{align}
      {1over k!} &= {1over k!}left(1 - {1over k} + {1over k}right) \
      &le {1over k!}left(1 + {1over k-1} - {1over k}right) \
      &= {1over k!}left({kover k -1} - {1over k}right) \
      &= {1over (k - 1)(k-1)!} - {1over kcdot k!}
      end{align}
      $$



      By telescoping we obtain:
      $$
      left|sum_{k=n+1}^m {1over k!}right| le left|{1over ncdot n!} - {1over mcdot m!}right|
      $$



      Given $m>n$ and $n,m in Bbb N$:
      $$
      left|{1over ncdot n!} - {1over mcdot m!}right| le left|1over ncdot n!right| = {1 over ncdot n!}
      $$



      Applying the limit to $|x_m - x_n|$ one may obtain:
      $$
      0 le lim_{ntoinfty}|x_{n+p} - x_n| le lim_{ntoinfty} {1over ncdot n!} = 0
      $$



      So squeezing $|x_m - x_n|$ gives:
      $$
      lim_{ntoinfty}|x_{n+p} - x_n| = 0, pin Bbb N
      $$



      Which would eventually mean:
      $$
      |x_m - x_n| < epsilon
      $$



      However I'm not sure how to find the index $N_epsilon$ from which the inequality becomes true since the expression for the upper bound involves a factorial.



      If we now put $x = x_0 in Bbb R$:
      $$
      0 le lim_{ntoinfty}|x_{n+p} - x_n| le lim_{ntoinfty} {x_0over ncdot n!} = 0
      $$
      Which doesn't influence the value of the limit.




      There are three questions in my mind:




      1. Is the overall reasoning valid?

      2. Is it possible to find a closed form of $N(epsilon)$, such that $n, m > N_epsilon implies |x_n - x_m| < epsilon$?

      3. Should I consider two cases for $xge 0$ and $x<0$




      Thank you!






      calculus sequences-and-series limits proof-verification cauchy-sequences




      calculus sequences-and-series limits proof-verification cauchy-sequences






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 11:29







      roman

















      asked Dec 26 '18 at 11:22









      romanroman

      2,28421224




      2,28421224






















          1 Answer
          1






          active

          oldest

          votes


















          6












          $begingroup$

          You can simply use ratio test to show that the given series converges. Any sequence $S$ is converging if
          $$r=lim_{nto infty}frac{a_{n+1}}{a_n}<1$$
          For this problem,
          $$r=frac{x^{n+1}}{(n+1)!}frac{n!}{x^n}$$
          $$=frac{x}{n+1}$$
          $r$ goes to $0$ as $ntoinfty$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh, I completely forgot of the ration test. Thank you!
            $endgroup$
            – roman
            Dec 26 '18 at 13:01










          • $begingroup$
            You're welcome ;)
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 13:02











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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          You can simply use ratio test to show that the given series converges. Any sequence $S$ is converging if
          $$r=lim_{nto infty}frac{a_{n+1}}{a_n}<1$$
          For this problem,
          $$r=frac{x^{n+1}}{(n+1)!}frac{n!}{x^n}$$
          $$=frac{x}{n+1}$$
          $r$ goes to $0$ as $ntoinfty$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh, I completely forgot of the ration test. Thank you!
            $endgroup$
            – roman
            Dec 26 '18 at 13:01










          • $begingroup$
            You're welcome ;)
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 13:02
















          6












          $begingroup$

          You can simply use ratio test to show that the given series converges. Any sequence $S$ is converging if
          $$r=lim_{nto infty}frac{a_{n+1}}{a_n}<1$$
          For this problem,
          $$r=frac{x^{n+1}}{(n+1)!}frac{n!}{x^n}$$
          $$=frac{x}{n+1}$$
          $r$ goes to $0$ as $ntoinfty$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh, I completely forgot of the ration test. Thank you!
            $endgroup$
            – roman
            Dec 26 '18 at 13:01










          • $begingroup$
            You're welcome ;)
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 13:02














          6












          6








          6





          $begingroup$

          You can simply use ratio test to show that the given series converges. Any sequence $S$ is converging if
          $$r=lim_{nto infty}frac{a_{n+1}}{a_n}<1$$
          For this problem,
          $$r=frac{x^{n+1}}{(n+1)!}frac{n!}{x^n}$$
          $$=frac{x}{n+1}$$
          $r$ goes to $0$ as $ntoinfty$.






          share|cite|improve this answer









          $endgroup$



          You can simply use ratio test to show that the given series converges. Any sequence $S$ is converging if
          $$r=lim_{nto infty}frac{a_{n+1}}{a_n}<1$$
          For this problem,
          $$r=frac{x^{n+1}}{(n+1)!}frac{n!}{x^n}$$
          $$=frac{x}{n+1}$$
          $r$ goes to $0$ as $ntoinfty$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 26 '18 at 11:33









          Ankit KumarAnkit Kumar

          1,494221




          1,494221












          • $begingroup$
            Oh, I completely forgot of the ration test. Thank you!
            $endgroup$
            – roman
            Dec 26 '18 at 13:01










          • $begingroup$
            You're welcome ;)
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 13:02


















          • $begingroup$
            Oh, I completely forgot of the ration test. Thank you!
            $endgroup$
            – roman
            Dec 26 '18 at 13:01










          • $begingroup$
            You're welcome ;)
            $endgroup$
            – Ankit Kumar
            Dec 26 '18 at 13:02
















          $begingroup$
          Oh, I completely forgot of the ration test. Thank you!
          $endgroup$
          – roman
          Dec 26 '18 at 13:01




          $begingroup$
          Oh, I completely forgot of the ration test. Thank you!
          $endgroup$
          – roman
          Dec 26 '18 at 13:01












          $begingroup$
          You're welcome ;)
          $endgroup$
          – Ankit Kumar
          Dec 26 '18 at 13:02




          $begingroup$
          You're welcome ;)
          $endgroup$
          – Ankit Kumar
          Dec 26 '18 at 13:02


















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