5
$\begingroup$

I am having a particularly difficult time working with the sequence of functions given by:

$$f_n(x)=\dfrac{1+nx^2}{(1+x)^n}.$$

I am working only on $[0,1]$. It is clear that these are measurable functions which go to $0$ pointwise, (though $f_n(0)=1\;\forall n\in\mathbb{N}$) and I would like to use either the Dominated or Monotone Convergence Theorems to show that:

$$\lim\limits_{n\to\infty}\int\limits_0^1f_n(x)=0.$$

I know it is true based on the context in which I received the problem, but it is proving to be a tough nut to crack. I'm usually good at these.

A little noodling around on Desmos shows that the sequence is in fact dominated on $[0,1]$ by any constant function greater than 1, and that for $n\geq 3$ it is decreasing on $[0,1]$. For $n=1$ or $n=2$, it has a minimum between $0$ and $1$. But these things are, for the most part, incredibly tedious to show rigorously.

This was a question on a prior preliminary exam in my department, and was written to take a few minutes at most.

Is there a quicker way? All these details have taken me almost an hour to write out!

  • 0
    What is the question? It seems that you have found a dominated function.2017-01-05
  • 0
    The question is in the last line. I'm looking for a quick and rigorous solution. My current one takes a very long time to write out, and the problem was not written to take so long.2017-01-05
  • 0
    Oh, I thought you had given the answer on your own since you said "A little noodling around on Desmos shows that the sequence is in fact dominated on $[0,1]$ by any constant function greater than $1$". What was the calculation? Maybe your calculation is not that complicated?2017-01-05
  • 0
    Desmos is just an online graphing calculator. I was doing it visually, which is helpful but clearly not allowed on the exam, and certainly not rigorous in any sense.2017-01-05

1 Answers 1

5

By the Bernoulli's inequality, for $x\in[0,1]$ $$ \frac{1+nx^2}{(1+x)^n}\leq\frac{1+nx^2}{1+nx}\leq 1 $$ which gives the dominated function.

  • 1
    Perfect! This is why I use Math.SE. Never heard of this, but now that I see it, it is a perfect solution to the problem. These little nuggets are so valuable. Thanks so much!2017-01-05