Find $\displaystyle \lim_{n\to\infty}\frac{n}{2^{n}}\int_{0}^{1}{x^{2n }\left( x^{2}+1 \right)^{n}\,dx}.$
Let $\displaystyle I_{m,n} = \int^{1}_{0} x^{m} (x^{2} + 1)^{n} \; \mathrm{d}x$
Then by using integration by parts,
$\displaystyle I_{m,n} = \frac{2^{m}}{m+1} - \frac{2n}{m+1} I_{m+2,n-1}$
Setting m = 2n, and using this repeatedly gives an expression for the integral:
$\displaystyle I_{2n,n} = 2^{n}\left[\frac{1}{2n+1} - \frac{n}{(2n+1)(2n+3)} + \cdots + (-1)^{n} \frac{n!}{(2n+1)(2n+3)\cdots(4n+1)} \right]$
Where the general term is $\displaystyle (-1)^{r} \frac{n(n-1) \cdots (n+1-r)}{(2n+1)(2n+3) \cdots (2n + 2r + 1)}$
So $\displaystyle \lim_{n \to \infty} \frac{n}{2^{n}} \int^{1}_{0} x^{2n}(x^{2}+1)^{n}\;\mathrm{d}x = \frac{1}{2}-\frac{1}{4}+\frac{1}{8} - \cdots = \frac{1}{3}$
That's right.
I saw this problem in another forum yesterday. Here's another solution:
Let $\displaystyle \lambda =\int_{0}^{1}{x^{2n}\left( x^{2}+1 \right)^{n}\,dx}.$ Now let $\displaystyle x\to x^2$ and the integral becomes $\displaystyle \lambda =\frac{1}{2}\int_{0}^{1}{(1+x)^{n}x^{n-1/2}\,dx},$ let $\displaystyle x\to1-x$ and the integral becomes $\displaystyle \lambda =2^{n-1}\int_{0}^{1}{\bigg\{ 1-\frac{x}{2} \bigg\}^{n}(1-x)^{n-1/2}\,dx},$ let $\displaystyle x\to nx$ and the integral becomes $\displaystyle \lambda =\frac{2^{n-1}}{n}\int_{0}^{n}{\bigg\{ 1-\frac{x}{2n} \bigg\}^{n}\bigg\{ 1-\frac{x}{n} \bigg\}^{n-1/2}\,dx}.$
Now the limit becomes $\displaystyle \lim_{n\to\infty}\frac12\int_{0}^{n}{\bigg\{ 1-\frac{x}{2n} \bigg\}^{n}\bigg\{ 1-\frac{x}{n} \bigg\}^{n-1/2}\,dx}.$ Finally, by Dominated Convergence Theorem the limit equals $\displaystyle \frac{1}{2}\int_{0}^{\infty }{e^{-x/2}\cdot e^{-x}\,dx}=\frac{1}{2}\int_{0}^{\infty }{e^{-3x/2}\,dx}=\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}.$
Hey Krizalid,
Dominated convergence theorem!!
What is this Dominated convergence theorem? (The wiki article is greek to me, with "measure spaces" )
Can you,please, state the conditions and the theorem and illustrate an interesting example case where this theorem is powerful?
Thank you,
Isomorphism
It basically means if $\displaystyle f_n\to f$ you cannot pass the limit through because of possible pointwise convergence.
If $\displaystyle f_n\to f$ uniformly it is okay, but Krizalid also passes the limit to the $\displaystyle n$ term getting $\displaystyle \infty$ in the upper limit.