Prove $\sum^{\infty}_{n=1} \frac{a_{n+1}-a_{n}}{a_{n}}=\infty$ for an increasing sequence $a_n$ of positive integers The $a_n$'s are integers, positive, and increasing: $0< a_1 < a_2 < \cdots$, the problem asks us to prove that:
$$
\sum^{\infty}_{n=1} \frac{a_{n+1}-a_{n}}{a_{n}}=\infty
$$ 
While I have checked this results for several series like $a_n = n$, $a_n = n^2$, $a_n = n^p$, or $a_n = p^n$ type stuff, I don't know how to prove this general result. A hint is appreciated. Thanks dudes!
 A: We have $$\sum_{i=1}^{n} \frac{a_{i+1}-{a_i}}{a_i} \geq \sum_{i=1}^{n}\int_{a_i}^{a_{i+1}}\frac{1}{x}\rm{d}x=\ln\left(\frac{a_{n+1}}{a_1} \right )$$
taking $n\rightarrow \infty$ we get the result.
A: I never thought I'd answer a question asked by a superhero! I would advise Mr. Iron Man to use the following well-known theorem:

Let $\{x_i\}$ be a sequence of positive real numbers. Then the product
$$\prod_{i=1}^\infty (1+x_i)$$
converges if and only if the series
$$\sum_{i=1}^\infty x_i$$
converges.

In the present case case, notice that
$$1+\frac{a_{i+1}-a_i}{a_i} = \frac{a_{i+1}}{a_i}$$
so the partial products of the infinite product telescope, to give $a_{n+1}/a_1$, which tends to $+\infty$ by assumption. Therefore, the series $\sum \frac{a_{i+1}-a_i}{a_i}$ diverges. 
Remark Your series is analogous to the integral $$\int_0^\infty df/f$$ where $f$ is a positive function. Of course, this integral equals $\varinjlim_{x \to \infty} \log (f(x)/f(0))$, which is $+ \infty$ if $f \to \infty$.
