Take $f(x)$ to be an arbitrary function on $x\in\mathbb{N}$. Assume $\lim_{x\rightarrow \infty} \frac{f(x)}{x}=c \not = 0$ therefore $f(x)$ diverges? Assume $\lim_{x\rightarrow \infty} \frac{f(x)}{x}=c \not = 0$.
To show that $f(x)$ necessarily diverges, I was told it follows from the limit test. That is as if to say:
$\sum_{x \in [1,\infty)\subset\mathbb{Z}} \frac{f(x)}{x}=\infty$ implies that $f(x)$ will also diverge. Is there a proof or some intuition about this one may offer?
 A: It doesn't really seem to make much difference that we restrict
$x \in \Bbb Z, \tag 1$
for if
$\displaystyle \lim_{x \to \infty} \dfrac{f(x)}{x} = c \ne 0, \tag 2$
then by definition, given any
$\epsilon > 0, \tag 3$
there exists
$M > 0 \tag 4$
such that
$x > M \Longrightarrow \left \vert \dfrac{f(x)}{x} - c \right \vert < \epsilon; \tag 5$
that is,
$-\epsilon < \dfrac{f(x)}{x} - c < \epsilon, \tag 6$
or
$-\epsilon x < f(x) - cx < \epsilon x, \tag 7$
whence
$(c - \epsilon)x < f(x) < (c + \epsilon)x; \tag 8$
now if 
$c > 0 \tag 9$
we choose
$\epsilon < \dfrac{c}{2}; \tag{10}$
then
$c - \epsilon > \dfrac{c}{2} > 0, \tag{11}$
and it follows that
$\displaystyle \lim_{x \to \infty} \left ( c - \epsilon \right ) x = \infty, \tag{12}$
whence (8) implies
$\displaystyle \lim_{x \to \infty} f(x) = \infty \tag{13}$
as well; on the other hand, with
$c < 0, \tag{14}$
we choose 
$0 < \epsilon < -\dfrac{c}{2}; \tag{15}$
now
$c + \epsilon < -\dfrac{c}{2}, \tag{16}$
whence
$\lim_{x \to \infty} (c + \epsilon)x = -\infty, \tag{17}$
and (8) yields
$\lim_{x \to \infty} f(x) = -\infty; \tag{18}$
we see that $f(x)$ diverges for all $c \ne 0$.
A: Well, since $$\lim_{x\to\infty}\frac{f(x)}x=c\neq 0,$$ then there is some $N$ such that, for all $x\ge N,$ we have $$\left|\frac{f(x)}x-c\right|<|c|.$$ From this, it follows that there is some $d>0$ such that


*

*$\frac{f(x)}x\ge d$ whenever $x\ge N,$ or

*$\frac{f(x)}x\le -d$ whenever $x\ge N.$
Consequently, we have for all $n>N$ that $$\left|\sum_{x\in[N,n]\cap\Bbb Z}\frac{f(x)}x\right|=\sum_{x\in[N,n]\cap\Bbb Z}\left|\frac{f(x)}x\right|=\sum_{x\in[N,n]\cap\Bbb Z}d=(n-N)d.$$
Thus, $$\left|\sum_{x\in[N,\infty)\cap\Bbb Z}\frac{f(x)}x\right|=\lim_{n\to\infty}\left|\sum_{x\in[N,n]\cap\Bbb Z}\frac{f(x)}x\right|=\lim_{n\to\infty}(n-N)d=\infty.$$
Since one of the facts 1 or 2 must hold, this means that either $$\sum_{x\in[N,\infty)\cap\Bbb Z}\frac{f(x)}x=\infty$$ or $$\sum_{x\in[N,\infty)\cap\Bbb Z}\frac{f(x)}x=-\infty.$$ In either case, $$\sum_{x\in[1,\infty)\cap\Bbb Z}\frac{f(x)}x$$ diverges, as desired.
A: Since $c\neq 0$ and $f(x) /x\to c$ as $x\to\infty $, it follows that $f(x) /x$ (and therefore $f(x) $) maintains same sign as $c$ and $|f(x) | > (|c|/2)x$ for large $x$. It follows that $f(x) \to\infty $ or $f(x) \to-\infty $ as $x\to\infty $ depending on whether $c>0$ or $c<0$.
