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Let $\left(\frac1{(2k+1)!}\right)$ be an infinite sequence. I want to show the following limit

$\lim \limits_{k \to \infty}{\frac{1}{(2k+1)!}}=0.$

Below is my proof. Please check it, I'm not confident about my ability.

For all nonnegative integer of k, we have

$0<(2k+1)\le(2k+1)!$

$\frac{1}{2k+1}\ge\frac{1}{(2k+1)!}>0.$

As $k\to\infty, \frac{1}{2k+1}\to0.$ Thus, as $k\to\infty$, $\frac1{(2k+1)!}\to0$

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  • $\begingroup$ Alternatively, you can try $\frac{e^x-e^{-x}}2=??$ $\endgroup$ Jul 27, 2013 at 16:44
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    $\begingroup$ Looks like a great application of the squeeze theorem to me! $\endgroup$ Jul 27, 2013 at 16:48
  • $\begingroup$ Your proof is correct=) $\endgroup$ Jul 27, 2013 at 16:49
  • $\begingroup$ Proof is correct. You may also try with $\frac{1}{(2k+1)!} \le \frac{1}{2^{2k}}$ $\endgroup$
    – Supriyo
    Jul 27, 2013 at 16:50
  • $\begingroup$ Your proof is perfectly good if we can take it as known that $\lim_{k\to\infty}\frac{1}{2k+1}=0$, and if we can take the Squeezing Theorem as known. If we need a formal $\epsilon$-$N$ argument, both of the above facts must be proved. So whether a proof is adequate is context-dependent. $\endgroup$ Jul 27, 2013 at 16:55

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As mentioned in the comments, your proof is great. :)

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