Convergence of series $\sum_{n=1}^{\infty} \frac{g'(n)}{g(n)}$ Suppose that $g, g', (g')^2 - gg''$ are all positive real valued functions that exist on the domain $[1, \infty).$ Prove that $$\sum_{n=1}^{\infty} \frac{g'(n)}{g(n)}$$ converges if and only if $$\displaystyle{\lim_{x \to \infty} g(x) < \infty} .$$
If $\sum_{n=1}^{\infty} \frac{g'(n)}{g(n)}$ converges then $ \lim_{n \to \infty} \frac{g'(n)}{g(n)} = 0.$ How can I obtain the result from here? Any hints?
 A: HINT
$$\sum_{n=1}^{\infty} \frac{g'(n)}{g(n)}\approx \lim_{a\to \infty}\int_1^a \frac{g'(x)}{g(x)}dx=\lim_{a\to \infty}[\ln (g(x))]_1^a$$
A: Like what @mathworker & @user siad:
Notice that $(g')^2-gg''>0$ is equivalent to $(\frac{g}{g'})'<0$. So it is a strict decreasing function. Now by $g,g'>0$, Lower Riemann Sum is Right Riemann Sum, Hence we have:
$$\sum_{n=1}^{\infty}\frac{g'(n)}{g(n)}=\frac{g'(1)}{g(1)}+\sum_{n=2}^{\infty}\frac{g'(n)}{g(n)}<\frac{g'(1)}{g(1)}+\int_{1}^{\infty}\frac{g'(n)}{g(n)}=\frac{g'(1)}{g(1)}+\ln g(\infty)-\ln g(1)$$
Also Upper Riemann Sum is left Riemann Sum, Hence we have:
$$\sum_{n=1}^{\infty}\frac{g'(n)}{g(n)}>\int_{1}^{\infty}\frac{g'(n)}{g(n)}=\ln g(\infty)-\ln g(1)$$
By mixing two inequalities:
$$\ln g(\infty)-\ln g(1)<\sum_{n=1}^{\infty}\frac{g'(n)}{g(n)}<\frac{g'(1)}{g(1)}+\ln g(\infty)-\ln g(1)$$
$$\sum_{n=2}^{\infty}\frac{g'(n)}{g(n)}+\ln g(1)<\ln g(\infty)<\sum_{n=1}^{\infty}\frac{g'(n)}{g(n)}+\ln g(1)$$
Since other terms are finite, convergence of series is equivalent to $\ln g(\infty)$ that is equivalent to $g(\infty)$.
