Proving Pointwise Convergence of a Sequence $\{f_n\}$ with a given $f_n$

Question

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?

Answer 1

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 $\ln(x_0^2)$): $$ \ln(1+y)\stackrel{!}{=} \ln(1)+\ln(y) $$

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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).

Answer 2

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.

Answer 3

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.

Answer 4

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 $