Limit of divided integrals Prove that: $$\lim\limits_{m\to \infty} \frac{\int\limits_0^{\pi/2}(\sin(x))^{2m} dx}{\int\limits_0^{\pi/2}(\sin(x))^{2m+1} dx} = 1$$
I had tried to get it from inequalities like $$(\sin(x))^{2m+1} \le (\sin(x))^{2m} \le (\sin(x))^{2m-1}$$
But I always hit a wall in that regard (I integrate the sin functions and from there keep on going but there it ends).
 A: You were on the right track. Let $I_n = \int_{0}^{\pi/2}\sin(x)^n\,dx$. Then $\{I_n\}_{n\geq 0}$ is clearly a decreasing sequence convergent to zero, and by integration by parts
$$ I_{2m+2} = \frac{2m+1}{2m+2}\,I_{2m}$$
In particular
$$ \frac{I_{2m}}{I_{2m+1}} \geq \frac{I_{2m+1}}{I_{2m+1}} = 1 $$
and
$$ \frac{I_{2m}}{I_{2m+1}} \leq \frac{I_{2m}}{I_{2m+2}} = 1+\frac{1}{2m+1}, $$
so the claim follows by squeezing. You may also prove $\frac{I_{2m}}{I_{2m+1}}\sim 1+\frac{2}{8m+3}$ through essentially the same technique.
A: You can calculate the integrals easily by using the first reduction formula at http://www.vias.org/calculus/07_trigonometric_functions_05_03.html.  Notice that the term outside the integral becomes $0$ when the limits are integration are substituted.
A: Just perform the integrals.
The numerator is:
$$\frac{\sqrt{\pi } \Gamma \left(m+\frac{1}{2}\right)}{2 \Gamma (m+1)}$$
and the denominator is
$$\frac{\sqrt{\pi } \Gamma (m+1)}{2 \Gamma \left(m+\frac{3}{2}\right)}$$
so the limit arises directly.
A: If you have a look at Table of Integrals, Series, and Products (Seventh Edition)  by I.S. Gradshteyn and I.M. Ryzhik, you should find that
$$\int\limits_0^{\pi/2}(\sin(x))^{2m}\, dx=\frac{(2m-1)!! }{(2m)!! } \frac \pi 2=\frac{\sqrt{\pi }\, \Gamma \left(m+\frac{1}{2}\right)}{2\, \Gamma (m+1)}$$
$$\int\limits_0^{\pi/2}(\sin(x))^{2m+1}\, dx=\frac{(2m)!! }{(2m+1)!! }=\frac{\sqrt{\pi }\, \Gamma (m+1)}{2 \,\Gamma \left(m+\frac{3}{2}\right)}$$ (formulae FI II 151). This makes
$$I_m=\frac{\int\limits_0^{\pi/2}(\sin(x))^{2m}\, dx}{\int\limits_0^{\pi/2}(\sin(x))^{2m+1}\, dx}= \frac{\Gamma \left(m+\frac{1}{2}\right) \Gamma \left(m+\frac{3}{2}\right)}{\Gamma
   (m+1)^2}$$ Now, take logarithms, use Stirling approximation and continue with Taylor series for large values of $m$ to get
$$\log(I_m)=\frac{1}{4 m}-\frac{1}{8 m^2}+O\left(\frac{1}{m^3}\right)$$
$$I_m=e^{\log(I_m)}=1+\frac{1}{4 m}-\frac{3}{32 m^2}+O\left(\frac{1}{m^3}\right)$$ which, for sure, shows the limit but also how it is approached.
As shown in the table below, the approximation is quite good even for small values of $m$.
$$\left(
\begin{array}{ccc}
 m & \text{exact} &  \text{approximation} \\
 1 & 1.17810 & 1.15625 \\
 2 & 1.10447 & 1.10156 \\
 3 & 1.07379 & 1.07292 \\
 4 & 1.05701 & 1.05664 \\
 5 & 1.04644 & 1.04625 \\
 6 & 1.03917 & 1.03906 \\
 7 & 1.03387 & 1.03380 \\
 8 & 1.02983 & 1.02979 \\
 9 & 1.02665 & 1.02662 \\
 10 & 1.02409 & 1.02406
\end{array}
\right)$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{\pi/2}\sin^{n}\pars{x}\,\dd x & =
\int_{0}^{\pi/2}\cos^{n}\pars{x}\,\dd x =
\int_{0}^{\pi/2}\exp\pars{n\ln\pars{\cos\pars{x}}}\,\dd x
\\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}
\int_{0}^{\infty}\expo{-nx^{2}/2}
\pars{1 - {x^{4} \over 12}}\,\dd x
\quad\pars{~Laplace\ Method~}
\\[5mm] & = \root{\pi \over 2}{1 \over \root{n}} -
{1 \over 4}\root{\pi \over 2}{1 \over n^{5/2}}
\end{align}

$$
\implies \lim_{m \to \infty}{\ds{\int_{0}^{\pi/2}\sin^{2m}\pars{x}\,\dd x} \over
\ds{\int_{0}^{\pi/2}\sin^{2m + 1}\pars{x}\,\dd x}} =
\lim_{m \to \infty}\root{2m + 1 \over 2m} = \bbx{\Large 1}
$$
