Lebesgue's Dominated Convergence Theorem problem 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
 A: 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.
A: 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 $1$.
A: $$
\lim_{n \rightarrow \infty}\int_{0}^{\infty}\frac{n}{(1+y)^n(ny)^{1/n}}dy = \lim_{n \rightarrow \infty}\int_{0}^{\infty}  (1+\frac{x}{n})^{-n}x^{-\frac{1}{n}}dx
 $$
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 
$$
\int \liminf_{n \rightarrow \infty} f_{n}du = \int \limsup_{n \rightarrow \infty} f_{n}du = \int_{0}^{\infty}e^{-x}dx = 1
$$
