# $f:[1,∞) \to \mathbb R$ a uniformly continuous function and $A_n=f(n)$ is a series.

Prove the following:

$$f:[1,\infty)\to\mathbb{R}$$ a uniformly continuous function and $$A_n=f(n)$$ is a series. Prove that if limit $$A_n$$ in infinity is plus infinity, then limit $$f(x)$$ in infinity is plus infinity.

My attempt: I started with that $$f([x]) = f(n) = A_n$$, so I need to prove that $$f(x)-f([x])$$ converges to zero and so limit $$f(x)$$ in infinity is plus infinity, and I can see that it will somewhat similar to the proof of Cauchy's term/condition of functions, but I'm getting successful in proving the sentence

• If $f(x)=x$, then $f(x)-f([x])$ doesn't converge, if $f(x)=x^2$ then $f(x)-f([x])$ isn't bounded Jun 15, 2019 at 23:36
• @user11513173 Check the edit, please. Jun 16, 2019 at 2:06
• @Nosrati thanks for the edit Jun 16, 2019 at 6:57
• @Pedro I am sorry but I didn't get what you mean.. Jun 16, 2019 at 6:57
• @Luyw it means to round the number to the closest integer from the bottom Jun 16, 2019 at 15:08

Claim : there is a finite constant $$M$$ such that $$|f(x)-f(y)| \leq M$$ whenever $$|x-y| \leq 1$$. Indeed choose $$\delta$$ such that $$|f(x)-f(y)| <1$$ whenever $$|x-y| <\delta$$. Any interval of length $$1$$ can be covered by at most $$1+ [\frac 1 {\delta}]$$ intervals of length $$\delta$$. The claim follow from this and the fact that $$f(x) \to \infty$$ as $$x \to \infty$$ is now easy to prove.
More details: choose $$\delta$$ corresponding to $$\epsilon =1$$ in the definition of uniform continuity. Let $$x. Let $$N =[\frac 1 {\delta}]+1$$. (Here $$[t]$$ is the greatest integer less than or equal to $$t$$). Then $$N >\frac 1 {\delta}$$ Divide the interval $$[x,y]$$ into $$N$$ equal parts by points $$x_0,x_1,...,x_N$$. The $$x_i-x_{i-1}=\frac {y-x} N <\delta$$ provided $$y-x \leq 1$$. Hence $$|f(x_i)-f(x_{i-1})| <1$$ for each $$i$$. This gives $$|f(y)-f(x)| \leq \sum |f(x_i)-f(x_{i-1}| We have proved that $$|f(y)-f(x)| whenever $$|x-y| \leq 1$$. Once you have this you can finish the proof by noting that $$n \leq x \leq n+1$$ implies $$|f(x)-f(n)| so $$f(x) \geq f(n)-N$$ which can be made as large as we want because $$f(n) \to \infty$$.