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In answering this question regarding this limit:

$$\lim \limits_{n\to \infty }\sin^2 \left(\pi \sqrt{(n!)^2-(n!)}\right)$$

I started by stating that $\sqrt{(n!)^2-(n!)} \to n!$, so that

$$\begin{split} \lim \limits_{n\to \infty }\sin^2 \left(\pi \sqrt{(n!)^2-(n!)}\right) &\stackrel{?}{=} \lim \limits_{n\to \infty }\sin^2 \left(\pi n!\right) \\ &= \lim \limits_{2k\to \infty }\sin^2 \left(\pi 2k\right) \\ &= 0. \end{split} $$

However, that is wrong. I can see that from the different other approaches that the solution is really 1 (which I verified). What makes my approach invalid?

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Barry
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  • I think the first answer here http://math.stackexchange.com/questions/46065/lim-n-to-infty-cos-pi-sqrtn2-n-second-battle?lq=1 is relevant – Nescio Jun 09 '15 at 13:52
  • $\sqrt{(n!)^2-(n!)} \to n!$ doesn't mean anything: the value of a limit doesn't depend on $n$. – Bernard Jun 09 '15 at 13:56

1 Answers1

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It is not true that $\sqrt{(n!)^2-(n!)}\to n!$. It is impossible for $f(n)$ to have a limit that depends on $n$ - the limit is a constant, if it exists.

It is true that $\sqrt{(n!)^2-(n!)}\sim n!$, where $f(n)\sim g(n)$ means $\frac{f(n)}{g(n)}\to 1$, but that is different from saying that $f(n)\to g(n)$.

The limit you want to use is probably: $$\lim_{m\to\infty}\left(\sqrt{m^2-m}-m \right)= -\frac{1}{2}\tag{1}$$

where you'll obviously replace $m=n!$.

You can prove (1) by showing that:

$$\sqrt{m^2-m}-m = -\frac{m}{\sqrt{m^2-m}+m}=-\frac{1}{\sqrt{1+\frac{1}{m}}+1}$$

Thomas Andrews
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