1

The function sequence $f_n=nx^{n-1}$ tends to zero for all $x \in [0,1)$. Why is the above limit not equal to $0$? Can you show it with Lebesgue's convergence theorems?

Gary
  • 31,845
  • 2
    what must you check for before applying Lebesgue's DOMINATED convergence theorem? (anyway, to calculate the integral, integration by parts is a very good idea... look at the numerator. And then check whether dominated convergence can be applied) – peek-a-boo Apr 01 '21 at 17:54
  • 1
    Put $x = t^{1/n}$ and use the dominated convergence theorem. – Gary Apr 01 '21 at 17:54
  • Doesn't this sequence of functions converge uniformly on $[0,1]$? – Anon Apr 01 '21 at 17:55
  • No, it does not even converge on [0,1]. –  Apr 01 '21 at 17:56

1 Answers1

2

Since $nx^{n-1}$ is the derivative of $x^n$, it suggests we first integrate by parts:

$$\int_0^1 \frac{nx^{n-1}}{1+x}\, dx = \frac{1}{2} + \int_0^1 \frac{x^n}{(1+x)^2}\,dx.$$

Now you can use the dominated (or even just the monotone) convergence theorem to conclude that $$\lim_{n \to \infty} \int_0^1 \frac{x^n}{(1+x)^2} \, dx = 0.$$

Unit
  • 7,601
  • 1
    I would suggest you also to explain why the DCT is not applicable to the original integral. – user Apr 01 '21 at 18:21