# A limit without Taylor series or l'Hôpital's rule $\lim_{n\to\infty}\prod_{k=1}^{n}\cos \frac{k}{n\sqrt{n}}$

Computing without Taylor series or l'Hôpital's rule

$$\lim_{n\to\infty}\prod_{k=1}^{n}\cos \frac{k}{n\sqrt{n}}$$

What options would I have here? Thanks!

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The best option is to prove Taylor series formula and use this answer –  no identity Feb 24 '13 at 22:41
@Norbert: Teacher says that we need no Taylor here. –  Chris's sis Feb 24 '13 at 22:47
Did the teacher suggest anything, apart from saying what not to do? –  Pedro Tamaroff Feb 24 '13 at 22:56
@PeterTamaroff: no clue from his side. All I know is that he is always right. –  Chris's sis Feb 24 '13 at 22:58
Is using the fact that $\sin(x) \geq x-\frac{x^3}{6}$and $\log(1+x)\geq x-\frac{x^2}{2}$ legal? –  Ishan Banerjee Feb 25 '13 at 13:32

The following tries to use only "basic" inequalities, avoiding L'Hopital and Taylor.

The inequality $$\tag1 e^x\ge 1+x\qquad x\in\mathbb R$$ should be well-known and after taking logarithms immediately leads to $$\tag2 \ln(1+x)\le x\qquad x>-1$$ and after taking reciprocals $$\tag3 e^{-x}\le \frac1{1+x}\qquad x>-1$$ Substituting $-\frac x{1+x}$ for $x$ in $(3)$ and taking logarithms, we have $$\tag4 \frac x{1+x}\le \ln(1+x)\qquad x>-1.$$ For $0<x<\frac\pi2$ we have $$\ln\cos x=\ln\sqrt{1-\sin^2x}=\frac12\ln(1-\sin^2x)$$ and by $(2)$ and $(4)$ $$\tag 5 -\frac12\tan^2x=\frac12 \frac{-\sin^2x}{1-\sin^2x}\le \ln\cos x\le -\frac12\sin^2x.$$ Since $0\le \sin x \le x\le \tan x$ for $0\le x<\frac\pi2$ and the quotient of the upper and the lower bound in $(5)$ is just $\cos^2x$ we find for $0<x\le y<\frac\pi2$ (using $\cos^2y\le \cos^2x$) $$-\frac12 x^2\cdot\frac1{\cos^2y}\le \ln\cos x\le -\frac12 x^2\cdot \cos^2y.$$ Letting (for $n\ge1$) $y=\frac1{\sqrt n}$, $x=\frac k{n\sqrt n}$ with $1\le k\le n$, this gives $$-\frac{k^2}{2n^3}\cdot\frac{1}{\cos^2\frac1{\sqrt n}}\le \ln\cos \frac k{n\sqrt n}\le -\frac{k^2}{2n^3}\cdot\cos^2\frac1{\sqrt n}.$$ Using $1^2+2^2+\ldots + n^2=\frac{n(n+1)(2n+1)}{6}$ we find by summation $$-\frac{n(n+1)(2n+1)}{12n^3}\cdot\frac{1}{\cos^2\frac1{\sqrt n}}\le\sum_{k=1}^n\ln\cos\frac k{n\sqrt n}\le-\frac{n(n+1)(2n+1)}{12n^3}\cdot\cos^2\frac1{\sqrt n}$$ and by taking the limit $$\sum_{k=1}^\infty\ln\cos\frac k{n\sqrt n}=-\frac16$$ and ultimately $$\prod_{k=1}^\infty\cos\frac k{n\sqrt n}=e^{-\frac16}.$$

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I think it is $\exp(-1/6)$, but this makes sense. –  Pedro Tamaroff Feb 24 '13 at 23:37
Oops, I see I have $h^2$ in $(5)$ where there should be $\sqrt h$. Not only does this change th elimit to invoke l'Hopital, but it also makes the error estimate $(2)$ vs. $(1)$ wrong (though that could be mended). I'll have to rethink. –  Hagen von Eitzen Feb 25 '13 at 11:21
@HagenvonEitzen: your style of solving problems is somewhat different from the "usual" styles. I like to see that. Consequently, I definitely choose your answer. –  Chris's sis Feb 25 '13 at 16:48

The best way (I think) to solve this problem is to use $\ln(1-x)=-x+O(x^2)$ and $\sin x=x+O(x^3)$. In fact, \begin{eqnarray*} \lim_{n\to\infty}\ln\prod_{k=1}^n\cos\frac{k}{n\sqrt{n}}&=&\lim_{n\to\infty}\sum_{k=1}^n\ln\cos\frac{k}{n\sqrt{n}}=\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^n\ln\cos^2\frac{k}{n\sqrt{n}}\\ &=&\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^n\ln(1-\sin^2\frac{k}{n\sqrt{n}})\\ &=&-\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^n\left[\left(\frac{k}{n\sqrt{n}}\right)^2+O\left(\left(\frac{k}{n\sqrt{n}}\right)^4\right)\right]\\ &=&-\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^n\frac{k^2}{n^3}=-\frac{1}{6}. \end{eqnarray*} Hence $$\lim_{n\to\infty}\prod_{k=1}^n\cos\frac{k}{n\sqrt{n}}=e^{-\frac{1}{6}}.$$

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I am just going to prove that the product converges. To get the exact value, you need more advanced stuff.

Start with $|\cos(x)-1| \le x^2/2$. This can be proved from $\cos(2x) = \cos^2(x)-\sin^2(x) = 1 - 2 \sin^2(x)$, so $|\cos(2x)-1| = 2|\sin^2(x)|$. Since $|\sin(x)| \le x$ for all $x$, $|\cos(2x)-1| \le 2x^2$ or $|\cos(x)-1| \le x^2/2$.

Since $|\cos(x)| \le 1 + x^2/2$, $|\cos(\frac{k}{n\sqrt{n}})| \le 1 + \frac{k^2}{2n^3}$. Since $\sum_{k=1}^n \frac{k^2}{2n^3} = \frac1{2n^3} \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{12n^3} < 1$ for $n > 1$, the product converges (since $\prod (1+a_n) < \exp(\sum a_n)$ if $a_n > 0$).

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