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Consider the sequence $\{a_n\}$ defined by

$$ a_n = \sqrt{n}\cdot \int_{0}^{\pi} \left( \frac{1 + \cos t}{2} \right)^n dt.$$

An exercise in Rudin, Real and Complex Analysis, requires showing that this sequence is convergent to a real number $a$, with $ a > 0$. I don't have any idea of how to prove this. I only obtained the following estimation

$$ \begin{align*} \int_{0}^{\pi} \left( \frac{1 + \cos t}{2} \right)^n dt &= 2 \int_{0}^{\frac{\pi}{2}} \left( 1 - \sin^2 t \right)^n dt \\ &> 2 \int_{0}^{\frac{1}{\sqrt{n}}} (1 - t^2)^n dt \\ &> 2 \int_{0}^{\frac{1}{\sqrt{n}}} (1 - n t^2) dt \\ & = \frac{4}{3 \sqrt{n}}, \end{align*}$$

which shows that $ a_n > \frac{4}{3}$.

Thank you very much in advance for any help.

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5 Answers 5

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Edited. The original answer is appended.

Put $$c_m:=\int_0^{\pi/2} \cos^m s\ ds\qquad(m\geq0)\ .$$ Then $c_1=1$, $\> c_2={\pi\over4}$, and an easy partial integration shows that $$c_m={m-1\over m} c_{m-2}\qquad(m\geq2)\ .\tag{1}$$ Now $$a_n:=\sqrt{\mathstrut n}\int_0^\pi\left({1+\cos t\over2}\right)^n\ dt=2\sqrt{\mathstrut n}\int_0^{\pi/2}\cos^{2n}s\ ds$$ and therefore $$a^2_n=4 n\> c_{2n}^2\ .$$ I claim that $$a_n^2={c_{2n}\over c_{2n-1}}\pi\qquad(n\geq1)\ .\tag{2}$$ Proof by induction: $(2)$ is true for $n=1$. Assume that it is true for $n-1$. Using $(1)$ we then have $$\eqalign{a_n^2 &= {n\over n-1}\left({c_{2n}\over c_{2n-2}}\right)^2 a_{n-1}^2 = {n\over n-1}{2n-1\over 2n}{c_{2n}\over c_{2n-2}}{c_{2n-2}\over c_{2n-3}}\pi\cr &={n\over n-1}{2n-1\over 2n}{2n-2\over 2n-1}{c_{2n}\over c_{2n-1}}\pi={c_{2n}\over c_{2n-1}}\pi\ .\qquad\square\cr}$$ Since $c_m\leq c_{m-1}\leq c_{m-2}$ it follows easily from $(1)$ that $$\lim_{m\to\infty}{c_m\over c_{m-1}}=1\ .$$ From $(2)$ we therefore conclude that $\lim_{n\to\infty} a_n=\sqrt{\mathstrut \pi}$.

Original answer: Let $$b_n:=\int_0^\pi\left({1+\cos t\over2}\right)^n\ dt=2\int_0^{\pi/2}\cos^{2n}s\ ds\ .$$ Then $$b_n =2\int_0^{\pi/2}\cos^{2n-2}s(1-\sin^2 s)\ ds=b_{n-1}-2\int_0^{\pi/2}\bigl(\cos^{2n-1}s\>\sin s\bigr)\cdot\sin s\ ds\ .$$ We integrate the last integral by parts and obtain $$b_n=b_{n-1}-{1\over 2n-1} b_n\ .$$ Therefore the $b_n$ satisfy the recursion $$b_n={2n-1\over 2n}\>b_{n-1}\qquad(n\geq1)\ ,$$ and the $a_n=\sqrt{\mathstrut n}\>b_n$ satisfy the recursion $$a_n^2={(2n-1)^2\over 2n\>(2n-2)}\>a^2_{n-1}\qquad(n\geq2)\ .$$ As $a_1={\pi\over2}$ we therefore have $$a_n^2=2\cdot\left({1\over2}\cdot{3\over2}\cdot{3\over4}\cdot{5\over4}\cdot{5\over6}\cdots{2n-1\over2n-2}{2n-1\over2n}\right)\cdot{\pi^2\over4}\qquad(n\geq2)\ .$$ In the large parenthesis the reciprocal of Wallis' product appears. It follows that this parenthesis converges to ${2\over\pi}$, and we obtain $$\lim_{n\to\infty}a_n=\sqrt{\mathstrut \pi}\ .$$

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  • $\begingroup$ This is very, very nice! +1 $\endgroup$
    – Pedro
    Commented Jul 16, 2013 at 1:02
  • $\begingroup$ @PeterTamaroff I upvoted the answer (+1), but this is just a high school approch with a well-defined pathway. I'm not sure what you refer at when you say "very, very". $\endgroup$ Commented Jul 20, 2013 at 9:46
  • $\begingroup$ @PeterTamaroff en.wikipedia.org/wiki/Wallis_product $\endgroup$ Commented Jul 20, 2013 at 9:48
  • $\begingroup$ @Chris'swisesister "This is just a high school approach..." Are you serious? I find it very, very stupid to underestimate things like that. This answer is clear and to the point. The integral is just another way to encode Wallis' result, and the answer doesn't hide it. $\endgroup$
    – Pedro
    Commented Jul 20, 2013 at 16:40
  • $\begingroup$ @PeterTamaroff "integration by parts" is a very powerful tool, but I cannot say it's a "very very (many times) nice too" every time I see it because it's just an usual tool. This case above is the same. It's a well-known case of integral that is connected to Wallis product and where integration by parts leads to some recurrence, a well-well known way. The way is definitely nice, but you refer to it as if you saw it for the first time in your life. $\endgroup$ Commented Jul 20, 2013 at 19:30
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$$ \begin{align} \lim_{n\to\infty}\sqrt{n}\int_0^\pi\left(\frac{1+\cos(t)}{2}\right)^n\,\mathrm{d}t &=\lim_{n\to\infty}\sqrt{n}\int_0^\pi\cos^{2n}(t/2)\,\mathrm{d}t\\ &=\lim_{n\to\infty}2\sqrt{n}\int_0^{\pi/2}\cos^{2n}(t)\,\mathrm{d}t\\ &=\lim_{n\to\infty}2\sqrt{n}\int_0^{n^{-1/3}}\left(1-\sin^2(t)\right)^n\,\mathrm{d}t\\ &=\lim_{n\to\infty}2\sqrt{n}\int_0^{n^{-1/3}}\left(1-t^2+\color{#C00000}{O}\big(t^4\big)\right)^n\,\mathrm{d}t\\ &=\lim_{n\to\infty}2\int_0^{n^{1/6}}\left(1-t^2/n+\color{#C00000}{O}\big(t^4/n^2\big)\right)^n\,\mathrm{d}t\\ &=\lim_{n\to\infty}2\int_0^{n^{1/6}}e^{-t^2+\color{#00A000}{O}(t^4/n)}\,\mathrm{d}t\\ &=2\int_0^\infty e^{-t^2}\,\mathrm{d}t\\[6pt] &=\sqrt\pi \end{align} $$ Where, for $n\ge3$, the red $O$ has constant $1/3$ and the green $O$ has constant $2$.

Note that $$ \begin{align} \lim_{n\to\infty}\sqrt{n}\int_{n^{-1/3}}^{\pi/2}\left(1-\sin^2(t)\right)^n\,\mathrm{d}t &\le\lim_{n\to\infty}\frac\pi2\sqrt{n}\left(1-\frac4{\pi^2}n^{-2/3}\right)^n\\ &\le\lim_{n\to\infty}\frac\pi2\sqrt{n}\,e^{-4n^{1/3}/\pi^2}\\[4pt] &=0 \end{align} $$


Motivation

As Didier points out, this approach is less elegant than those that make use of certain trigonometric identities, and perhaps Rudin posed this problem using trigonometric functions because one can use trigonometric identities to get a more elegant solution. However, I had computed the integral of $\cos^n(t)$ too many times recently, and I wanted to single out key sufficient features of $f$ that allow the limit $$ \lim_{n\to\infty}\sqrt{n}\int_0^\pi f(t)^n\,\mathrm{d}t=\sqrt\pi $$ those being

  1. $f(t)=1-t^2+O(t^4)$

  2. $f(t)\le1-kt^2$ for some $k\gt0$

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  • $\begingroup$ And of course the previous-to-last equal sign is where everything happens (so much so that, with no further justification, it could cause violent reactions from a susceptible teacher...). $\endgroup$
    – Did
    Commented Jul 15, 2013 at 10:48
  • $\begingroup$ @Did: yes, I noticed that and I am working on it. $\endgroup$
    – robjohn
    Commented Jul 15, 2013 at 10:49
  • $\begingroup$ (+1) although the equality $1-\sin^2(t)=1-t^2+O(t^4)$ is valid only on a neighborhood of $0$. $\endgroup$
    – user63181
    Commented Jul 15, 2013 at 10:58
  • $\begingroup$ @SamiBenRomdhane: It is certainly valid for all $t$ since $1-\sin^2(t)\le1$. $\endgroup$
    – robjohn
    Commented Jul 15, 2013 at 11:01
  • $\begingroup$ @Did: hopefully, the violence can be avoided now. $\endgroup$
    – robjohn
    Commented Jul 15, 2013 at 11:20
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Related problem. One can evaluate the integral using the beta function

$$ \beta(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2u-1}(\cos\theta)^{2v-1}\,d\theta, \qquad \mathrm{Re}(u)>0,\ \mathrm{Re}(v)>0. $$

First, we use the identity $1+\cos(x)=2\cos^2(x/2)$

$$ a_n=n^{1/2}\int_{0}^{\pi}\left(\frac{1+\cos(x)}{2}\right)^n dx = n^{1/2}\int_{0}^{\pi}{\cos^{2n}(x/2)} dx. $$

Then using the substitution $y=\frac{x}{2}$, we have

$$ a_n= 2 n^{1/2}\int_{0}^{\pi/2}{\cos^{n/2}(y)} dy= \sqrt {\pi }{\frac {\sqrt {n}\,\Gamma \left( n+1/2 \right)}{\Gamma \left( n+1 \right) }}.$$

Taking the limit as $n\to \infty$, we get

$$ \lim_{n\to \infty} a_n = \sqrt{\pi}. $$

Note that, you can use Stirling approximation for $n!=\Gamma(n+1)\sim \left(\frac{n}{e}\right)^n\sqrt{ \pi n}. $ to evaluate the limit.

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  • $\begingroup$ Maybe $\frac{\Gamma(n+\tfrac{1}{2})}{\Gamma(n+1)} = \sqrt{\pi}\frac{(2n)!}{(n!)^2\,4^n}$ is helpful too. $\endgroup$
    – Nikolaj-K
    Commented Jul 15, 2013 at 11:43
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Note that $a_n=\int\limits_0^\infty u_n(t)\mathrm dt$, where $$ u_n(t)=2^{-n}(1+\cos(t/\sqrt{n}))^n\,\mathbf 1_{0\leqslant t\leqslant\pi\sqrt{n}}. $$ It happens that $u_n\to u$ pointwise (can you show this?), where $$ u(t)=\mathrm e^{-t^2/4}, $$ hence a natural conjecture is that $a_n\to a$, where $$ a=\int_0^\infty u(t)\mathrm dt=\sqrt\pi. $$ A tool to make sure this convergence happens is Lebesgue dominated convergence theorem, which requires to find some integrable $v$ such that $|u_n|\leqslant v$ for every $n$. It happens that $v=u$ fits the bill.

To see why, note that $\cos(2t)+1=2\cos^2(t)$ hence $|u_n|\leqslant u$ for every $n$ as soon as, for every $x$ in $(0,\pi/2)$, $\cos(x)\leqslant\mathrm e^{-x^2/2}$. Any idea to show this last (classical) inequality?

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  • $\begingroup$ I don't know how to prove the pointwise convergence to $e^{\frac{-t^2}{4}}$ nor the inequality you quoted. $\endgroup$ Commented Jul 15, 2013 at 11:27
  • $\begingroup$ Do you know the expansion of $\cos(t)$ around $t=0$? (Two terms suffice...) $\endgroup$
    – Did
    Commented Jul 15, 2013 at 23:50
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    $\begingroup$ Well, apparently, 7 hours ago you were suddenly quite able to fill the gaps of this solution, weren't you? $\endgroup$
    – Did
    Commented Jul 15, 2013 at 23:52
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(too long to be a comment)

The integrals can be calculated exactly:

$$a_n=\frac{n^{1/2}}{2^n} \int_0^\pi(1+\cos t)^n dt. $$ For starters, we can use Newton's binomial: $$a_n=\frac{n^{1/2}}{2^n}\int_0^\pi \sum_{k=0}^n \binom{n}{k}\cos^ktdt. $$ Use the even-ness of the cosine around $t=\pi$: $$a_n=\frac{n^{1/2}}{2^{n+1}}\sum_{k=0}^n \binom{n}{k} \int_0^{2\pi} \cos^k tdt $$

The integrals $\int_0^{2\pi} \cos^k t dt$ can be calculated using Euler's formula $$\int_0^{2\pi} \cos^k t dt=\begin{cases} 0 & k=2m+1 \\ \frac{2\pi}{2^{2m}} \binom{2m}{m} & k=2m \end{cases}. $$ Therefore the summation is only needed for even indices $k=2m$, and $$a_n=\frac{n^{1/2}}{2^{n+1}} \sum_{m=0}^{\lfloor{\frac{n}{2}} \rfloor} \binom{n}{2m} \frac{2\pi}{2^{2m}} \binom{2m}{m} $$

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