Uniform convergence of a series

WaterPoloGoat
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Homework Statement



We know that f is uniformly continuous.

For each n in N, we define fn(x)=f(x+1/n) (for all x in R).

Show that fn converges uniformly to f.

Homework Equations



http://en.wikipedia.org/wiki/Uniform_convergence

The Attempt at a Solution



I know that as n approaches infinity, that fn(x)=f(x), which implies that fn converges to f.

I'm currently trying to apply the fact that if fn is uniformly convergent, then

limn->infinity Sup {fn(x): x in R}=0.

But I keep getting stuck on the fact that there's an function in the definition of fn i.e., fn(x)=f(x+1/n). Is there a way to work with it?
 
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It's actually pretty straightforward. Given \epsilon > 0, for each x there is a \delta>0 such that |f(x+h)-f(x)|<\epsilon whenever |h|<\delta. Hence, for all n > \delta ^{-1} we have |f_n(x)-f(x)|<\epsilon.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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