$\displaystyle\lim_{x\to 0} {(\frac{f(x)}{x})} = 2$. Prove that the series $a_n=f(1)+f(\frac{1}{2})+...+(\frac{1}{n})$ is diverging to infinity I have a feeling this has something to do with the harmonic series. I feel like I need to find a step that does something like $f(x)=2x$ but I think this is not allowed.
 A: Given any $\varepsilon>0$, since $\lim_{x\rightarrow 0}\frac{f(x)}{x}=2$ we can choose $N$ large enough so that $\forall x<\frac{1}{N}$:$-\varepsilon<\frac{f(x)}{x}-2<\varepsilon$ $$\rightarrow\sum_{n=N}^{N+k}f(\frac{1}{n})>(2-\varepsilon)\sum_{n=N}^{N+k}\frac{1}{n}$$
for all $k>0$.
Now you just need to make $k$ large enough that the LHS go to infinity.
A: If
$\lim_{x \to 0} \frac{f(x)}{x}
=2
$,
then,
for some
$c > 0$,
$0 < x < c
\implies
\frac{f(x)}{x}
> 1
$
or
$f(x) > x$
or
$f(1/n)
> 1/n
$
for $n > 1/c
$.
Therefore,
if
$n > \frac{2}{c}
$,
$\sum_{k=1}^n f(1/k)
=\sum_{k \le 1/c} f(1/k)+\sum_{k>1/c}^{k \le n} f(1/k)
\gt\sum_{k>1/c}^{k \le n} 1/k
\gt \ln(n)-\ln(1/c)
\to \infty
$
as
$n \to \infty$.
A: You are correct. You may want to be more specific in the proof below (ie, apply more accurately the definition of the limit). My 2 cents: since the limit of $f(x)/x$ is 2 then, for exits $n_0$ such that for all $n > n_0$ we have $f(x)/x > 1, x = 1/n $ (in other words, when the argument $x$ get close to $0$ then $f(x)/x$ is in a vicinity of the limit $2$, and is bigger than $1$. Now: $a_n = a_{n0} + f(\frac{1}{n_0+1}) +  f(\frac{1}{n_0+2}) +\dots$ satisfies $a_n >= a_{n0} + \frac{1}{n_0+1}+\frac{1}{n_0+2}+\dots$, which is unbound (related to the harmonic series you mention). I hope this is somewhat clear...
A: $\forall\epsilon>0, \exists\delta>0 s.t. \forall x,  |x|<\delta \implies |\frac{f(x)}{x}-2|<\epsilon$
Set $x = \frac{1}{n}$ and we deal with positive n's, then we have $\forall n, n>\frac{1}{\delta} \implies 2-\epsilon < \frac{f(\frac{1}{n})}{\frac{1}{n}}<2+\epsilon$
Choosing $\epsilon<2$, we obtain $f(\frac{1}{n}) > 0$
Now you can use comparison rule.
Compare $a_n = \sum_{k=0}^n f(\frac{1}{k})$  with $b_n = \sum_{k=0}^n \frac{1}{k}$
We know $b_n$ is divergent.
$lim_{k \to \infty} \frac{f(\frac{1}{k})}{\frac{1}{k}}$ = $lim_{x \to 0} \frac{f(x)}{x} = 2$
By the rule $a_n$ is divergent.
