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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.

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    Try this [approach](http://math.stackexchange.com/questions/370023/how-to-prove-a-sequence-of-a-function-converges-uniformly/370071#370071).2013-06-07

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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. $

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    @LordFarin: unfortunately $\left(\cdots\right)^{1/2n}$ is not mathematically correct, so I reverted rather than add unsightly parentheses.2013-06-07