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My attempt: We need to show that $|xn-xm|<e$ for all $m,n \geq H(e) $ By archimedian property we can always have an H such that He>1 and $(1/H)<e$ we chose n and m such that m=(n+p) where p is a natural number . $|1+\frac{1}{(2)!}+.. +\frac{1}{(n)}! -1-\frac{1}{(2)!}..-\frac{1}{(n+p)!}| = |\frac{1}{(n+1)!}+\frac{1}{(n+2)!}...+\frac{1}{(n+p)!}| < |\frac{n}{(n+1)!}| = \frac{1}{((n-1)!(n+1))}<\frac{1}{(n+1)}<\frac{1}{n}$ now we know that $n\geq H$ so $1/n<1/H<e$ there fore the above criteria holds and it's a cauchy sequence . Is my attempt correct?

Antimony
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  • I think finally it should be independent of $n$ – MANI Dec 05 '19 at 05:26
  • I didn't get it. Can you be a bit more precise – Antimony Dec 05 '19 at 05:29
  • You can't chose m, n because m,n should be arbitrary and m,n=>n0 for a n0 – PRASTYA SUSANTO Dec 05 '19 at 05:36
  • There are not just $n$ terms, there are $p-n$ terms and your $p$ should be free not dependent upon $n$ – MANI Dec 05 '19 at 05:36
  • But isn’t this the sequence of partial sums for $e-1$? Why not use the Ratio Test to see that the series is convergent? If you wanted to prove the claim from first principles, then you could compare with the geometric series $\sum_0^\infty1/2^n$ to conclude that the partial sums are bounded above. Use the fact that your sequence is nondecreasing to conclude that it’s convergent. – Lubin Dec 05 '19 at 05:54
  • Shouldn't it be smaller than $|p/(n+1)!|$? – kingW3 Dec 05 '19 at 12:08
  • take a look here for a different proof – Masacroso Dec 05 '19 at 14:17

2 Answers2

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We know that the difference between any two terms in the sequence is $|\frac{1}{n!}-\frac{1}{m!}|$. Suppose without loss of generality that $n\leq m$, then since the sequence is monotonically decreasing, this suggests

$|\frac{1}{n!}-\frac{1}{m!} |\leq |\frac{1}{n!}-\frac{1}{(n+1)!}|= |\frac{1}{n!}-\frac{1}{n!(n+1)}| = |\frac{1}{n!} (1-\frac{1}{n+1})|$

Thus $\forall \epsilon>0,$ let $N(\epsilon)$ satisfy $|\frac{N!}{N+1}|\geq \frac{1}{\epsilon}$ s.t. if $n,m\geq N(\epsilon)$, then $|\frac{1}{n!}-\frac{1}{m!}| \leq \epsilon$ so that it is a Cauchy sequence.

Adam
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  • This sequence is not decreasing? – MANI Dec 05 '19 at 05:53
  • Well if $a_n = \frac{1}{n!}$, and if $n\leq m$, then $n!\leq m!$ which implies $\frac{1}{n!}\geq \frac{1}{m!}$. Edit: To clarify, I mean the sequence itself (not the sequence of partial sums) – Adam Dec 05 '19 at 05:56
  • Ok, I was confused with the partial sum. – MANI Dec 05 '19 at 06:01
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$m=n+p;$

$|S_m-S_n|= \sum_{k=n+1}^{n+p}1/(k!)\lt$

$\dfrac{1}{(n+1)!}(1+ \dfrac{1}{n+2} +$

$\dfrac{1}{(n+3)(n+2)} +$

$\dfrac{1}{(n+2)(n+3)....(n+p)})\lt $

$\dfrac{1}{(n+1)!}(1+\dfrac{1}{n+2}+$

$ \dfrac{1}{(n+2)^2}+..\dfrac{1}{(n+2)^{p-1}}) \lt$

$ \dfrac{1}{(n+1)!}(1+(1/2)+(1/2)^2 +..(1/2)^{p-1}) \lt 2\dfrac{1}{(n+1)!}\lt 2/n$.

Given $\epsilon >0$;

Archimedean principle:

There is a $n_0 \ge \ 2/\epsilon$ s.t.

for $m\ge n\ge n_0$ :

$|S_m-S_n| \lt 2/n \le 2/n_0 <\epsilon$.

Used: $1+(1/2)+(1/2)^2.......=2$(Infinite geometric series)

Peter Szilas
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  • Isn't it similar to the way I have attempted .. m , n ,p are arbitrary elements so I really couldn't get where I went wrong .. can you just clarify .. – Antimony Dec 05 '19 at 13:59
  • smita.I try.Your second line, line below the "now" you have ......$|\dfrac{n}{(n+1)!}|$, it should be $|\dfrac{p}{(n+1)!}|$, there are p terms $\dfrac{1}{(n+1)}$, check it. And then : p can be arbitrarily large (m=n+p), so how to proceed to get $\dfrac{p}{(n+1)!} $ arbitrarily small.Your thoughts? – Peter Szilas Dec 05 '19 at 15:53