Does $f_{n}(x)=(1+x^{2n})^{1/2n}$ converge uniformly on $\mathbb{R}$.

Question

Does the sequence of functions defined by $f_{n}(x)=(1+x^{2n})^{1/2n}$ converge uniformly on $\mathbb{R}$.

For testing uniform convergence i know if the sequence $x_{n} = \sup \: \{ |f_{n}(x)-f(x) | : x \in \mathbb{R}\}$ converges to $0$ then $f_{n} \to f$ uniformly. But I don't know how to actually apply this result.

In other words, you should first figure out whether or not $f_n$ converges pointwise -- meaning whether or not $\lim f_n(x)$ makes sense for every $x \in \mathbb{R}$. If it does, then set $f(x) = \lim f_n(x)$, and see if you can apply the result you stated (regarding $x_n$). — Jesse Madnick, May 24 '12 at 05:11
Also notice that if $x\neq 0$, $f_n(x) = |x| (1+x^{-2n})^{\frac{1}{2n}}$. — copper.hat, May 24 '12 at 05:40
Agree with Didier. The fun begins with determining what $f$ is. Notice that (before you take the root) when $|x|<1$ the constant $1$ dominates, whereas when $|x|>1$ the term $x^{2n}$ dominates for large indices $n$. — Jyrki Lahtonen, May 24 '12 at 06:14
There's also the estimate $(1+|\theta|)^{\frac{1}{2n}} \leq 1+ \frac{|\theta|}{2n}$. — copper.hat, May 24 '12 at 06:42
Well, thanks to various commenters, all the pieces are now on the table, so to speak, and what is left is to put them together. Let us hope this moment of true pedagogy is not ruined by the appearance of a full-bolts-on solution. — Did, May 24 '12 at 07:24
@robjohn Apologies duly noted. However: Why did you post a full answer? — Did, May 24 '12 at 13:16
@Didier: oops! I didn't realize it was a homework problem (is it?). — robjohn, May 24 '12 at 13:24
@Josh: did you intend $\displaystyle(1+x^{2n})^{1/(2n)}$ (as everyone has been assuming)? — robjohn, May 24 '12 at 13:42
@robjohn Perhaps it's reasonable to undelete your answer after more than a year? — Lord_Farin, Jun 07 '13 at 14:57

robjohn · Answer 1 · 2013-06-07T16:31:24.953

Without loss of generality, assume $x\ge0$ and $n\ge1$.

For $x\ge1$, the formula for the sum of a geometric series yields $$ \begin{align} (1+x^{2n})^{\raise{2pt}{\large\frac{1}{2n}}}-x &=\frac{(1+x^{2n})-x^{2n}}{\sum\limits_{k=1}^{2n}(1+x^{2n})^{\raise{2pt}{\large\frac{k-1}{2n}x^{2n-k}}}}\\ &\le\frac{1}{2n}\tag{1} \end{align} $$ For $x\le1$, the Mean Value Theorem says $$ \begin{align} (1+x^{2n})^{\raise{2pt}{\large\frac{1}{2n}}}-1 &\le2^{\raise{2pt}{\large\frac{1}{2n}}}-1\\ &=e^\xi\left(\frac{1}{2n}\log(2)-0\right)\\ &\le\frac{\log(2)}{\sqrt{2}\,n}\tag{2} \end{align} $$ for some $\xi\in(0,\frac{1}{2n}\log(2))$

Estimates $(1)$ and $(2)$ guarantee uniform convergence to $$ f(x)=\left\{\begin{array}{} 1&\text{if }|x|\le1\\ |x|&\text{if }|x|>1 \end{array}\right. $$

@LordFarin: unfortunately $\left(\cdots\right)^{1/2n}$ is not mathematically correct, so I reverted rather than add unsightly parentheses. — robjohn, Jun 07 '13 at 16:28

Does $f_{n}(x)=(1+x^{2n})^{1/2n}$ converge uniformly on $\mathbb{R}$.

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