1

The task is not to evaluate these, but something else which you would come to know after reading to the end of the post.

$\newcommand{\u}[2]{\underbrace{#1}_{#2}} \newcommand{\i}[4]{\int_{#3}^{#4}\left(#1\right){\rm d}#2} \newcommand{\b}[1]{\left(#1\right)} \newcommand{\sb}[1]{\left[#1\right]}$

Let us define: $$f(n)=\i{x^{n-1}\sin\b{\frac{\pi x}2}}{x}{0}{1}\\ g(n)=\i{x^{n-1}\cos\b{\frac{\pi x}2}}{x}{0}{1}$$ Now I change the variable, $t=\pi x/2$: $$f(n)=\u{\b{\frac2\pi}^n}{\zeta}\u{\i{t^{n-1}\sin t}{t}0{\pi/2}}{F(n)}$$ Similiarly: $$g(n)=\u{\b{\frac2\pi}^n}{\zeta}\u{\i{t^{n-1}\cos t}{t}0{\pi/2}}{G(n)}$$ Now Let us integrate by parts: $$F(n)=\sb{t^{n-1}(-\cos t)}_0^{\pi/2}-\i{(n-1)t^{n-2}(-\cos t)}{t}0{\pi/2}$$ So: $$F(n)=(n-1)G(n-1)$$ Now I need to find these: $$\xi_1=\lim_{n\to\infty}{(n+1)f(n)}$$ And: $$\xi_2=\lim_{n\to\infty}\frac{(3n+1)f(n)}{(2n+1)^2g(n)}$$ The second one can be done as: $$\xi_2=\lim_{n\to\infty}\frac{(3n+1)\zeta (n-1)G(n-1)}{(2n+1)^2\zeta G(n)}$$ And I think that $\lim_{n\to\infty}G(n)=\lim_{n\to\infty}G(n)$ [No proof, intuitive] that $\xi_2=\frac34$

Questions:

  • Is $\xi_2$ correct?
  • What advice/hint would you give me to find $\xi_1$?
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1 Answers1

3

For (1),use this well known result $$\lim_{n\to\infty}n\int_{0}^{1}x^{n-1}f(x)dx=f(1).f\in C[0,1]$$ see 1

for $(2)$,we have general $$\lim_{n\to\infty}\dfrac{\int_{0}^{1}g(x)[f(x)]^{n+1}dx}{\int_{0}^{1}g(x)[f(x)]^ndx}=\max_{x\in[0,1]}\{f(x)\};f,g\in C[0,1],{\rm{and}}\; f,g>0$$ so let $g(x)=\cos{\frac{\pi x}{2}},f(x)=x$,we have $$\xi_{2}=\lim_{n\to\infty}\dfrac{(3n+1)(n-1)g(n-1)}{(2n+1)^2g(n)}=\dfrac{3}{4}\lim_{n\to\infty}\max_{x\in [0,1]}x=\dfrac{3}{4}$$

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