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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

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  • $\begingroup$ 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 $\endgroup$ Jun 15, 2019 at 23:36
  • $\begingroup$ @user11513173 Check the edit, please. $\endgroup$
    – Nosrati
    Jun 16, 2019 at 2:06
  • $\begingroup$ @Nosrati thanks for the edit $\endgroup$ Jun 16, 2019 at 6:57
  • $\begingroup$ @Pedro I am sorry but I didn't get what you mean.. $\endgroup$ Jun 16, 2019 at 6:57
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    $\begingroup$ @Luyw it means to round the number to the closest integer from the bottom $\endgroup$ Jun 16, 2019 at 15:08

1 Answer 1

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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 $x<y$. 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 $|f(y)-f(x)| \leq \sum |f(x_i)-f(x_{i-1}| <N$ We have proved that $|f(y)-f(x)| <N$ whenever $|x-y| \leq 1$. Once you have this you can finish the proof by noting that $n \leq x \leq n+1$ implies $|f(x)-f(n)| <N$ so $f(x) \geq f(n)-N$ which can be made as large as we want because $f(n) \to \infty$.

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  • $\begingroup$ I have given hints. Let me know if you need a detailed proof. $\endgroup$ Jun 15, 2019 at 23:45
  • $\begingroup$ Thanks for the hints, I would much appreciate it if you could elaborate more :) $\endgroup$ Jun 16, 2019 at 6:56
  • $\begingroup$ @user11513173 I have added more details. $\endgroup$ Jun 16, 2019 at 11:46
  • $\begingroup$ Omg thank you, you are such a life saver sir!! Absolute saint:) $\endgroup$ Jun 16, 2019 at 15:10

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