1
$\begingroup$

I am trying to prove pointwise convergence of a sequence $\{f_n\}$ given $f_n:[0,\infty)\rightarrow \mathbb{R}$ with

$$f_n(x)=\frac{x^n}{1+x^{2n}}$$

I started by delineating each possible value for x, so I have that

$$f(x)=\begin{cases}0 & \mathrm{if} \space 0\le x \lt 1\\ \frac12 & \mathrm{if} \space x=1 \\ 0 & \mathrm{if} \space x \gt 1 \end{cases}$$

Now to prove. So I set out to show that $\lim_{n\rightarrow\infty}\frac{x^n}{1+x^{2n}}=0 \quad \forall x\in[0,1)\cup(1,\infty)$.

Now $\mathit{let} \, ε\gt0$ be given, choose any $x_0\in[0,1)\cup(1,\infty)$. Find $n^*$ such that $|f_n(x)-f(x)|\ltε\quad\forall n\ge n^*$.

$$\Rightarrow f_n(x_0)=\frac{x_0^n}{1+x_0^{2n}}$$

Since $f(x_0)=0$, then we have:

$$\left\lvert\frac{x_0^n}{1+x_0^{2n}}-0\right\rvert=\left\lvert\frac{x_0^n}{1+x_0^{2n}}\right\rvert=\frac{x_0^n}{1+x_0^{2n}}\ltε$$

since $x_0$ is positive or zero. Taking the natural log of both sides gives:

$$\ln\left({\frac{x_0^n}{1+x_0^{2n}}}\right)\lt\ln{ε}\quad\Rightarrow\quad n\ln({x_0})-n\ln({x_0})=0\ltε$$

But the problem is that now I no longer have $n$ in my equation so I cannot generalize in terms of $n\ge n^*$. Where did I go wrong, or where could I take a different approach?

$\endgroup$

4 Answers 4

2
$\begingroup$

There are several issues in your "proof".

First of all, you don't need the redundant subscript in $x_0$, which makes your writing unnecessarily busy.

The first $\Rightarrow$ does not make sense: not even wrong. There is no implication there.

Second, the estimate you want is $$ \left|\frac{x^n}{1+x^{2n}}-0\right|<\epsilon\tag{0} $$

You seem to apply the following incorrect formula when taking logarithm (also with a missing factor of $2$ when taking $\ln(x_0^2)$): $$ \ln(1+y)\stackrel{!}{=} \ln(1)+\ln(y) $$


If $x\in[0,1)$, then $$ \lim_{n\to\infty}f_n(x)=\frac{\lim_n x^n}{1+\lim_n (x^2)^n}=\frac{0}{1+0}=0\tag{1} $$

If $x>1$, then $$ \lim_{n\to\infty}f_n(x)=\lim_n\frac{(\frac{1}{x})^n}{(\frac{1}{x^2})^n+1} =\frac{\lim_n(\frac{1}{x})^n}{\lim_n(\frac{1}{x^2})^n+1} =\frac{0}{0+1}=0\tag{2} $$

If you do want an $\epsilon$-$N$ proof, you can simply prove that $\lim_{n\to\infty} r^n=0$ for $0\le|r|<1$ and then apply the limit laws to get (1) and (2).

$\endgroup$
1
  • 1
    $\begingroup$ You're right, I overuse "=>" when writing proofs, which is something I have to change. You are also correct about the log; I mistakenly broke the fraction into two parts, which obviously does not work. Thanks for the input, I'll restart with your method in mind. Thanks as well for the note about the ε-N proof. $\endgroup$ Feb 20, 2021 at 22:44
1
$\begingroup$

Set $g(t)=\frac{t}{1+t^2}$. Note that $f_n(x)=g(x^n)$.

If $x>1$ is fixed, then $x^n\to\infty$, as $n\to\infty$, so $\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}g(x^n)=\lim_{t\to\infty}g(t)=0$. On the other hand, if $x\in[0,1)$ is fixed, then $x^n\to0$, as $n\to\infty$, so $\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}g(x^n)=\lim_{t\to0}g(t)=0$. Finally, for $x=1$ we have that $f_n(1)=1/2$ for all $n$, so $f_n(1)\to1/2$.

So the sequence $(f_n)$ converges pointwise to the piecewise-defined function you specified.

$\endgroup$
1
$\begingroup$

Note that $$0< \frac{x_0^n }{1+x_0^{2n} } <\frac{x_0^n }{x_0^{2n} } = x_0^{-n}$$ With the squeeze lemma the desired convergence follows for $x_0 >1$. For $x_0<1$ we simply use that the numerator converges to zero while the denominator converges to one.

Hope this helps you.

$\endgroup$
1
$\begingroup$

I think it would be better to treat both cases separately. First, for $x\in [0,1) $, you have :

$$\forall n\geq0, |f_n(x)|\leq \frac{x^n}{2} $$

which goes to $0$ at infinity since $ 0\leq x<1 $.

Then, for $x\in (1,\infty)$, you can use the fact that $x^{2n}+1\sim x^{2n} $

So $f_n(x)\sim\frac{x^n}{x^{2n}}=x^{-n}\to_{\infty} 0 $

$\endgroup$

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.