Show that if $f:\mathbb{R}\rightarrow\mathbb{R}$ is uniformly continuous, then sequence $f(x+1),f(x+\frac{1}{2}),f(x+\frac{1}{3}),\ldots$ Show that if $f:\mathbb{R}\rightarrow\mathbb{R}$ is uniformly continuous, then sequence $f(x+1),f(x+\frac{1}{2}),f(x+\frac{1}{3}),\ldots$ is uniformly convergent.

Let $g_n(x)=f(x+\frac{1}{n})$. A natural candidate for the uniform limit is $g(x)=f(x)$. Fix $\epsilon>0$. From uniform continuity, there is a $\delta>0$ such that $|f(x)-f(x+\frac{1}{n})|\le\epsilon$ whenever $\frac{1}{n}\le\delta$. Thus for any $\epsilon>0$ there is $\frac{1}{\delta}>0$
 such that $|f(x)-g_n(x)|\le\epsilon$ whenever $n>\frac{1}{\delta}$. This proves uniform convergence.
I would be very happy if someone checked it.
Also, I wonder if there is some trivial counterexample showing that assumption about uniform continuity is necessary.
 A: Your proof is correct except for the following detail : avoid writing things like "there is a $\frac{1}{\delta}>0$". Here you need to consider the "ceiling" of $\frac{1}{\delta}$, the smallest integer  above $\frac{1}{\delta}$, which is denoted by $\lceil \frac{1}{\delta} \rceil$.
So you should have written something like : "Thus for any $\epsilon>0$ there is a $n_0>0$ (namely, $n_0=\lceil \frac{1}{\delta} \rceil$)
 such that $|f(x)-g_n(x)|\le\epsilon$ whenever $n>n_0$."
In answer to your second question, $f(x)=e^{x}$ is a simple counterexample (you can check that $|g_n-f|$ is unbounded for any $n$, in fact it tends to $+\infty$ as $x\to\infty$).
A: Yes, your proof is perfect.  To prove uniform convergence you need to prove that for a fixed $\epsilon > 0$, then $|f(x) - g_n(x)| \le \epsilon$ for all $x$ and $n$ big enough; this is exactly what you did.
The classical example of a non-uniformly continuous function works as a counterexample: let $f(x) = e^x$, and let $g_n(x) = f(x + 1/n)$.Then if $(g_n)$ converged uniformly to $f$, there would be some $N \ge 0$ such that for all $n \ge N$, $|g_n(x) - f(x)| \le 1$ for all $x$. However, $|g_n(x) - f(x)| = e^x (e^{1/n} - 1)$, and you can always find $x > 0$ such that this is bigger than $1$ (say $x = 1+\ln(1/(e^{1/N}-1))$).
