Proving $e$ is irrational using its reciprocal and the inequality $0<e^-1-s_2n-1<frac1(2n)!$.

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Here is a small part of the proof found on Wikipedia:



enter image description here



I am trying to understand the inequality used for the proof of $e$ being irrational. The inequality is




$0<e^-1-s_2n-1<frac1(2n)!$




Is the middle part the error of the series for $e^-1$ after the $n$th term? Or is it for another term in the series? And what does $frac1(2n)!$ siginify? Does it have something to do with the neglected terms after the series is stopped?




taylor-expansion proof-explanation irrational-numbers




share|cite|improve this question























    up vote
    1
    down vote

    favorite












    Here is a small part of the proof found on Wikipedia:



    enter image description here



    I am trying to understand the inequality used for the proof of $e$ being irrational. The inequality is




    $0<e^-1-s_2n-1<frac1(2n)!$




    Is the middle part the error of the series for $e^-1$ after the $n$th term? Or is it for another term in the series? And what does $frac1(2n)!$ siginify? Does it have something to do with the neglected terms after the series is stopped?




    taylor-expansion proof-explanation irrational-numbers




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Here is a small part of the proof found on Wikipedia:



      enter image description here



      I am trying to understand the inequality used for the proof of $e$ being irrational. The inequality is




      $0<e^-1-s_2n-1<frac1(2n)!$




      Is the middle part the error of the series for $e^-1$ after the $n$th term? Or is it for another term in the series? And what does $frac1(2n)!$ siginify? Does it have something to do with the neglected terms after the series is stopped?




      taylor-expansion proof-explanation irrational-numbers




      share|cite|improve this question











      Here is a small part of the proof found on Wikipedia:



      enter image description here



      I am trying to understand the inequality used for the proof of $e$ being irrational. The inequality is




      $0<e^-1-s_2n-1<frac1(2n)!$




      Is the middle part the error of the series for $e^-1$ after the $n$th term? Or is it for another term in the series? And what does $frac1(2n)!$ siginify? Does it have something to do with the neglected terms after the series is stopped?





      taylor-expansion proof-explanation irrational-numbers





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      asked Jul 15 at 20:31









      numericalorange

      1,226110




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          1 Answer
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          This results from Leibniz' criterion for alternating series: the error for a convergent alternating series when truncated at the $n$-th term is no more than the next term in absolute value, and it has the same sign.






          share|cite|improve this answer





















          • Thanks so much for taking the time to reply. I understand now where $frac1(2n)!$ comes from!
            – numericalorange
            Jul 15 at 20:47










          • May I ask whether the last term in $s_2k-1$ is $-frac1(2k-1)!$?
            – numericalorange
            Jul 15 at 20:51






          • 1




            Yes,so the next (non-written) term in the series is $;frac1(2k)!$.
            – Bernard
            Jul 15 at 20:57










          • Thank you, that really helps my understanding. I am not sure whether I should ask this as a new question or ask it here, but later in the proof it says that $(2k-1)!s_2k-1$ is always an integer. Is there a proof on here or on a resource website that shows this fact or is it trivial?
            – numericalorange
            Jul 15 at 21:09






          • 1




            It' trivial, if you think of it: since the denominators are factorials, each of them is a divisor of those that follow, so the lcm of the denominators is the last one.
            – Bernard
            Jul 15 at 21:32










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          1 Answer
          1






          active

          oldest

          votes









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          active

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          votes








          up vote
          2
          down vote



          accepted










          This results from Leibniz' criterion for alternating series: the error for a convergent alternating series when truncated at the $n$-th term is no more than the next term in absolute value, and it has the same sign.






          share|cite|improve this answer





















          • Thanks so much for taking the time to reply. I understand now where $frac1(2n)!$ comes from!
            – numericalorange
            Jul 15 at 20:47










          • May I ask whether the last term in $s_2k-1$ is $-frac1(2k-1)!$?
            – numericalorange
            Jul 15 at 20:51






          • 1




            Yes,so the next (non-written) term in the series is $;frac1(2k)!$.
            – Bernard
            Jul 15 at 20:57










          • Thank you, that really helps my understanding. I am not sure whether I should ask this as a new question or ask it here, but later in the proof it says that $(2k-1)!s_2k-1$ is always an integer. Is there a proof on here or on a resource website that shows this fact or is it trivial?
            – numericalorange
            Jul 15 at 21:09






          • 1




            It' trivial, if you think of it: since the denominators are factorials, each of them is a divisor of those that follow, so the lcm of the denominators is the last one.
            – Bernard
            Jul 15 at 21:32














          up vote
          2
          down vote



          accepted










          This results from Leibniz' criterion for alternating series: the error for a convergent alternating series when truncated at the $n$-th term is no more than the next term in absolute value, and it has the same sign.






          share|cite|improve this answer





















          • Thanks so much for taking the time to reply. I understand now where $frac1(2n)!$ comes from!
            – numericalorange
            Jul 15 at 20:47










          • May I ask whether the last term in $s_2k-1$ is $-frac1(2k-1)!$?
            – numericalorange
            Jul 15 at 20:51






          • 1




            Yes,so the next (non-written) term in the series is $;frac1(2k)!$.
            – Bernard
            Jul 15 at 20:57










          • Thank you, that really helps my understanding. I am not sure whether I should ask this as a new question or ask it here, but later in the proof it says that $(2k-1)!s_2k-1$ is always an integer. Is there a proof on here or on a resource website that shows this fact or is it trivial?
            – numericalorange
            Jul 15 at 21:09






          • 1




            It' trivial, if you think of it: since the denominators are factorials, each of them is a divisor of those that follow, so the lcm of the denominators is the last one.
            – Bernard
            Jul 15 at 21:32












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          This results from Leibniz' criterion for alternating series: the error for a convergent alternating series when truncated at the $n$-th term is no more than the next term in absolute value, and it has the same sign.






          share|cite|improve this answer













          This results from Leibniz' criterion for alternating series: the error for a convergent alternating series when truncated at the $n$-th term is no more than the next term in absolute value, and it has the same sign.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 15 at 20:36









          Bernard

          110k635103




          110k635103











          • Thanks so much for taking the time to reply. I understand now where $frac1(2n)!$ comes from!
            – numericalorange
            Jul 15 at 20:47










          • May I ask whether the last term in $s_2k-1$ is $-frac1(2k-1)!$?
            – numericalorange
            Jul 15 at 20:51






          • 1




            Yes,so the next (non-written) term in the series is $;frac1(2k)!$.
            – Bernard
            Jul 15 at 20:57










          • Thank you, that really helps my understanding. I am not sure whether I should ask this as a new question or ask it here, but later in the proof it says that $(2k-1)!s_2k-1$ is always an integer. Is there a proof on here or on a resource website that shows this fact or is it trivial?
            – numericalorange
            Jul 15 at 21:09






          • 1




            It' trivial, if you think of it: since the denominators are factorials, each of them is a divisor of those that follow, so the lcm of the denominators is the last one.
            – Bernard
            Jul 15 at 21:32
















          • Thanks so much for taking the time to reply. I understand now where $frac1(2n)!$ comes from!
            – numericalorange
            Jul 15 at 20:47










          • May I ask whether the last term in $s_2k-1$ is $-frac1(2k-1)!$?
            – numericalorange
            Jul 15 at 20:51






          • 1




            Yes,so the next (non-written) term in the series is $;frac1(2k)!$.
            – Bernard
            Jul 15 at 20:57










          • Thank you, that really helps my understanding. I am not sure whether I should ask this as a new question or ask it here, but later in the proof it says that $(2k-1)!s_2k-1$ is always an integer. Is there a proof on here or on a resource website that shows this fact or is it trivial?
            – numericalorange
            Jul 15 at 21:09






          • 1




            It' trivial, if you think of it: since the denominators are factorials, each of them is a divisor of those that follow, so the lcm of the denominators is the last one.
            – Bernard
            Jul 15 at 21:32















          Thanks so much for taking the time to reply. I understand now where $frac1(2n)!$ comes from!
          – numericalorange
          Jul 15 at 20:47




          Thanks so much for taking the time to reply. I understand now where $frac1(2n)!$ comes from!
          – numericalorange
          Jul 15 at 20:47












          May I ask whether the last term in $s_2k-1$ is $-frac1(2k-1)!$?
          – numericalorange
          Jul 15 at 20:51




          May I ask whether the last term in $s_2k-1$ is $-frac1(2k-1)!$?
          – numericalorange
          Jul 15 at 20:51




          1




          1




          Yes,so the next (non-written) term in the series is $;frac1(2k)!$.
          – Bernard
          Jul 15 at 20:57




          Yes,so the next (non-written) term in the series is $;frac1(2k)!$.
          – Bernard
          Jul 15 at 20:57












          Thank you, that really helps my understanding. I am not sure whether I should ask this as a new question or ask it here, but later in the proof it says that $(2k-1)!s_2k-1$ is always an integer. Is there a proof on here or on a resource website that shows this fact or is it trivial?
          – numericalorange
          Jul 15 at 21:09




          Thank you, that really helps my understanding. I am not sure whether I should ask this as a new question or ask it here, but later in the proof it says that $(2k-1)!s_2k-1$ is always an integer. Is there a proof on here or on a resource website that shows this fact or is it trivial?
          – numericalorange
          Jul 15 at 21:09




          1




          1




          It' trivial, if you think of it: since the denominators are factorials, each of them is a divisor of those that follow, so the lcm of the denominators is the last one.
          – Bernard
          Jul 15 at 21:32




          It' trivial, if you think of it: since the denominators are factorials, each of them is a divisor of those that follow, so the lcm of the denominators is the last one.
          – Bernard
          Jul 15 at 21:32












           

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