convergence of $\sum_n x_n$ versus $\sum_n f(x_n)$ for differentiable $f$ Suppose $f(0) = 0$, $f$ is differentiable at 0, $f'(0) \ne 0$, and $x_n \rightarrow 0$. What can we say about (i) the convergence of $\sum_n x_n$ versus (ii) the convergence of $\sum_n f(x_n)$?
It seems that (i) implies (ii) and my reasoning is as follows. By the definition of derivative, $\lim_{x \rightarrow 0} \left( \frac{f(x)}{x} - f(0) \right) = 0$. Let $\epsilon = \left| \frac{f'(0)}{2} \right| > 0$. By the definition of limit, there is a $\delta > 0$ such that for all $x \in B(0, \delta)$, $\left| \frac{f(x)}{x} - f'(0) \right| < \epsilon$. Therefore for $x \in B(0, \delta)$, $\frac{1}{2}f'(0) x \le f(x) \le \frac{3}{2}f'(0) x$ and for some $N \in \mathbb{N}$, $x_n \in B(0, \delta)$ if $n > N$. So $\frac{1}{2}f'(0) \sum_{n > N} x_n = \sum_{n > N} \frac{1}{2}f'(0) x_n \le \sum_{n > N} f(x_n) \le \sum_{n > N} \frac{3}{2}f'(0) x_n = \frac{3}{2}f'(0) \sum_{n > N} x_n$.
It also seems that (ii) implies (i) by a similar argument. Specifically, $f$ must be strictly monotone in some open set $U$ containing 0 and we can apply the argument above to $(f\restriction_U)^{-1}$.
Is my reasoning right?


Specifically, $f$ must be strictly monotone in some open set $U$ containing $0$

No, it need not be. Unless you require continuous differentiability. Consider, for some $0 < c < \frac{1}{2}$, the function
$$f(x) = c\cdot x + \int_0^x \sin \tfrac{1}{t}\,dt.$$
$f$ is differentiable on all of $\mathbb{R}$, with $f'(0) = c\neq 0$, and $f'(x) = c + \sin \frac{1}{x}$ for $x\neq 0$. Then $f'$ changes sign infinitely often in every neighbourhood of $0$.
It is straightforward to see that an $f$ with your properties preserves absolute convergence, i.e. $\sum x_n$ converges absolutely if and only if $\sum f(x_n)$ converges absolutely, but for conditional convergence, even the harmless
$$f(x) = x + x^2$$
converts the conditionally convergent series
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{\sqrt{n}}$$
into the divergent series
$$\sum_{n=1}^\infty \left(\frac{(-1)^{n-1}}{\sqrt{n}} + \frac{1}{n}\right).$$
