0
$\begingroup$

I'm working through practice problems and I came along the following:

Evaluate $\lim_{n\rightarrow\infty}\int_0^n(1-\frac{x}{n})^n dx.$

I think this should work out via Dominated Convergence Theorem, but I can't seem to get it. I have been starting with $x\in(0,n)\Rightarrow 0<(1-\frac{x}{n})<1 \Rightarrow (1-\frac{x}{n})^n\leq(1-\frac{x}{n}).$ I just don't see how to use this. I can't get a bound for the integral that doesn't rely upon $n.$

Am I coming at it from the wrong angle? Any assistance is appreciated.

$\endgroup$
0
$\begingroup$

For sufficiently large $n$, we have $$ \int_0^n\left(1-\frac{x}{n}\right)^ndx=\frac{(x-n)(1-\frac{x}{n})^n}{n+1}\Bigg|_0^n=\frac{n}{n+1}. $$ What is the limit of that as $n$ goes to infinity?

$\endgroup$
3
$\begingroup$

Show that the following inequality is always satisfied: $$\left(1-\dfrac{x}n\right)^n \mathbb{1}_{x \in [0,n]} \leq e^{-x}$$ Then use Lebesgue dominated convergence theorem.

$\endgroup$
  • $\begingroup$ @downvoted Care to comment? $\endgroup$ – Leg May 20 '15 at 18:31
2
$\begingroup$

You can't use dominated convergence because of the $n$ in the upper limit. Why not just compute the integral directly?

$$\int_0^n \left(1-\frac{x}{n}\right)^n dx=\frac{n}{n+1}$$

So $$\lim\limits_{n\to\infty} \int_0^n \left(1-\frac{x}{n}\right)^n dx=1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.