1

I'm trying to prove that for every $n\in\mathbb{N} $ this inequality holds:

$0<\left|e-S_{n}\right|<\frac{3}{(n+1)!} $

S.T: $S_{n}=1+1+\frac{1}{2!}+\dots+\frac{1}{n!} $

By induction:

assuming for $n$ and proving for $n+1$:

$\left|e-S_{n+1}\right|=\left|e-S_{n}-\frac{1}{(n+1)!}\right|=\left|(e-S_{n})+(-\frac{1}{(n+1)!})\right|<|e-S_{n}|+\left|-\frac{1}{(n+1)!}\right|<\frac{3}{(n+1)!}+\left|-\frac{1}{(n+1)!}\right| $

but how can i continue from this stage?

edit note: i can't use the fact that: $lim_{n\rightarrow\infty}S_{n}=e $

thank you.

Amit
  • 139
  • This using of the triangle inequality doesn't seem me efficient because your estimation will be greater than $\frac{3}{(n+1)!}$. Instead of that use the following equality: $$e - S_n =\frac{1}{(n+1)!}+ \frac{1}{(n+2)!} + \frac{1}{(n+3)!}+, \ldots$$ – Leonhardt von M Apr 03 '15 at 09:12
  • first of all thank you for the comment, but i can't use the fact that: $lim_{n\rightarrow\infty}S_{n}=e $ – Amit Apr 03 '15 at 09:18
  • State in the question your definition of $e$ and what you are allowed to use, if there are limitations. – mickep Apr 03 '15 at 09:19
  • Okay, but please, define $e$. I usually define it as the limit of that sum XD There are a lot of possible ways of definition that number. – Leonhardt von M Apr 03 '15 at 09:22
  • Can you use the remainder form of Taylor's Theorem in your proof? – Paul Apr 03 '15 at 09:23
  • lol i know, but in this stage this limit is yet proved. just this one: $lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n}=e $ – Amit Apr 03 '15 at 09:23
  • yes, i can use the reminder of taylor in my proof! – Amit Apr 03 '15 at 09:24
  • Consider the $\lim\limits_{k\to \infty} (S_n - (1+\frac{1}{k})^k)$ and use the binomial theorem. You'll get my suggested equation. – Leonhardt von M Apr 03 '15 at 09:28
  • @Amit then there is a value a such that 0<a<1 for which the remainder term is $\frac{e^a}{(n+1)!}$. – Paul Apr 03 '15 at 09:39

1 Answers1

2

Hint: $$ \begin{align} (n+1)!\,(e-S_n) &=\sum_{k=1}^\infty\frac{(n+1)!}{(n+k)!}\\ &\le\sum_{k=1}^\infty\frac1{(n+2)^{k-1}}\\ &=\frac{n+2}{n+1} \end{align} $$

Doing this without the series representation seems difficult. We could use this answer to show that $$ \left[\lim_{n\to\infty}\left(1+\frac1n\right)^n\right]^x=\sum_{n=0}^\infty\frac{x^n}{n!} $$

Community
  • 1
robjohn
  • 345,667
  • This answer begs the question, imo...and also the OP said he can't use the series definition of $;e;$ . – Timbuc Apr 03 '15 at 10:10
  • 1
    Ah, I see that now. We could use this answer to show the series representation from the definition they have. I would be very interested in another approach using that definition. – robjohn Apr 03 '15 at 11:19
  • problem solved, with using the remainder form of Taylor's Theorem. thank you very much all! – Amit Apr 03 '15 at 11:34
  • 1
    @Amit Isn't "the Taylor Theorem" exactly the same as the series representation for $;e;$ ? – Timbuc Apr 03 '15 at 11:46
  • Generation of the Taylor series would need the definition of $e^x$ that says it is its own derivative. This also makes $e^x$ monotonic increasing. A bound on the remainder term then requires that e<3, which can also be obtained from the Taylor series (first 3 terms should do it probably). – Paul Apr 03 '15 at 17:01