Does the following series converge for the integral criterion for convergence? if $f>0$ is differenciable on $[0, \infty)$ and $f'$ decrease to zero, prove that the following series converges or diverges simultaneously
$$\sum_{n=1}^{\infty} f'(n), \quad  \sum_{n=1}^{\infty} \frac{f'(n)}{f(n)}$$
Use the integral criterion for convergence.
Because $f'$ es decrease to zero and is positive ( I'm not really sure about this fact, I thinks is positive because $f>0$) so for the integral criterion  $\sum_{n=1}^{\infty} f'(n)$ converges iff $\int_{1}^{\infty} f'(x)$ converge, for  the Fundametal Theorem of the calculus,  $$\int_{1}^{\infty} f'(x)= f(\infty)-f(1)$$
But I don't think is correct.
 A: We assume $f$ is positive and differentiable on $[0,\infty),$ with $f'$ decreasing and tending to $0$ as $x\to \infty.$ It follows that $f'\ge 0$ everywhere, which implies $f$ is increasing.
One direction in the equivalence is easy: Suppose $\sum f'(n)<\infty.$ Since $f$ is increasing, we have $f\ge f(0)$ everwhere. Thus
$$\sum_{n=1}^{\infty}\frac{f'(n)}{f(n)} \le \sum_{n=1}^{\infty}\frac{f'(n)}{f(0)} = \frac{1}{f(0)}\sum_{n=1}^{\infty}f'(n) <\infty.$$
In the other direction, suppose $\sum\dfrac{f'(n)}{f(n)}<\infty.$ Note that because $f'$ decreases and $f$ increases, $f'/f$ decreases. Thus by the integral test,
$$\int_0^\infty\frac{f'(x)}{f(x)}\,dx <\infty.$$
Since $f'/f = (\log f)',$ we have
$$\int_0^\infty(\log f)'(x)\,dx<\infty.$$
But the last integral equals $\log f\big | _0^\infty.$ It follows that $f$ is bounded on $[0,\infty)$ by a constant I'll call $f(\infty).$ We then see
$$\sum f'(n) = \sum f(n)\frac{f'(n)}{f(n)} \le \sum f(\infty)\frac{f'(n)}{f(n)} = f(\infty)\sum \frac{f'(n)}{f(n)}<\infty.$$
