# Proving that jte sequence $f_n(x)= (1+x^n)^{1/n}$ converges uniformly

Let's consider the sequence of functions $f_n : [0,2] \to \mathbb{R}$ defined by $f_n(x)= (1+x^n)^{1/n}$. I proved that this sequence converges pointwise to the function: $f(x) = 1$ if $0\le x \le 1$ and $f(x)=x$ otherwise. My problem is how can I prove that this sequence converges uniformly. (maybe I can prove that $f_n$ is a Cauchy sequence but it look complicated) Please help me )=

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what did you try? did you try proving it straight out of the definition? those functions on [0,2] are bounded, try to use that. Also, did you prove the series is monotone decreasing for a given X? Try to prove the general case (uniform convergence occurs in all such cases of monotone decreasing sequences of functions) – Benjamin Gruenbaum Oct 31 '12 at 7:12

• If $0\leqslant x\leqslant1$, then $1\leqslant f_n(x)\leqslant2^{1/n}\leqslant1+1/n$.
• If $x\geqslant1$, then $x\lt f(x)=x\cdot(1+x^{-n})^{1/n}\leqslant x\cdot(1+x^{-n}/n)\leqslant x+1/n$ since $n\geqslant1$.
This proves that $f(x)\leqslant f_n\leqslant f(x)+1/n$, hence $\|f_n-f\|_\infty\leqslant1/n\to0$. (The exact value being $\|f_n-f\|_\infty=2^{1/n}-1\sim\log2/n$, the order of this upper bound is the correct one.)
@Daniel Without them nothing guarantees the convergence to $1$, respectively to $x$. – Did Oct 31 '12 at 11:09