Lebesgue's Dominated Convergence Theorem problem

Question

I am having trouble using DCT for the following

Prove $$\lim_{n\to\infty}\int_0^\infty \frac{n}{(1+y)^n(ny)^\frac{1}{n}}dy = 1$$

I think most of the mass of the integral lies beneath $(e-1)/n$ but now dont believe this is true, so I was hoping to break up the integral into two parts, but that didn't seem to help. So I seem to be stuck at the moment.

Any help or suggestions would be greatly appreciated. It would be more helpful to me if you gave me a general suggestion rather than a specific one. Such as how to find the dominating function for integrals which become tidal waves as $n \rightarrow \infty$ or how to chose the point to split the integral into two bits.

Thanks

Answer 1

As I say in the comment, this is false as stated, but I am assuming the OP forgot $\lim_{n\rightarrow \infty}.$ If so, then for a fixed $n$ this is a standard beta-function integral, which is equal to $$\frac{n^{-1/n} \Gamma \left(2-\frac{1}{n}\right) \Gamma \left(n-2+\frac{1}{n}\right)}{\Gamma (n).}$$

This, in turn, goes to $0,$ not $1$ as $n$ goes to infinity, so either the claimed answer is wrong, or there is another typo somewhere.

Answer 2

I agree with Igor that the limit is $0$. You can still prove this claim using DCT though. You only need to worry about large values of $n$, so pick $N=3$ say, then when $y>1$ you can bound the function by $$\frac{y}{(1+y)^{3}}\leq\frac{1}{(1+y)^{2}}$$ which is integrable. On the other hand for $y<1$, you can bound by $$y^{1-\frac{1}{3}}$$ which is integrable on $[0,1]$. So that piecewise defined function has only one point of discontinuity, hence integrable, and DCT applies. However, if you exchange the limit with the integral you get $0$...not $1$.

Answer 3

$$ \lim_{n \rightarrow \infty}\int_{0}^{\infty}\frac{n}{(1+y)^n(ny)^{1/n}}dy = \lim_{n \rightarrow \infty}\int_{0}^{\infty} (1+\frac{x}{n})^{-n}x^{-\frac{1}{n}}dx $$ by substitute $y = \frac{x}{n}$.

Let $f_{n} = (1+\frac{x}{n})^{-n}x^{-\frac{1}{n}}$

Then, by DCT, it is clear that the above integral is $1$.

As a matter of fact, I don't know what is $g$ that dominating $f_{n}$.

I am not sure, but you can use Fatou's lemma for $\liminf$ and $\limsup$ using $$ \int \liminf_{n \rightarrow \infty} f_{n}du \leq \liminf_{n \rightarrow \infty} \int f_{n}du \leq \limsup_{n \rightarrow \infty} \int f_{n}du \leq \int \limsup_{n \rightarrow \infty} f_{n}du $$

where $$ \int \liminf_{n \rightarrow \infty} f_{n}du = \int \limsup_{n \rightarrow \infty} f_{n}du = \int_{0}^{\infty}e^{-x}dx = 1 $$