Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
...and dudettes! – Bruno Joyal Nov 13 '13 at 18:49
@tony:do not yet have the privilege of posting a comment, so putting this comment as an answer.Did you try out the sum for the situation when a(n+1)/a(n) = 1+(1/n^2)? – Sudhir Nov 14 '13 at 16:46
Dear Sudhir: in the question, the $a_n$'s are integers. – Bruno Joyal Nov 14 '13 at 18:27
Consider a partial sum: $\displaystyle{\large S_{N} \equiv \sum_{n = 1}^{N}{a_{n + 1} - a_{n} \over a_{n}} = \sum_{n = 1}^{N}{a_{n + 1} \over a_{n}} - N > 0}$. – Felix Marin Nov 15 '13 at 4:23
up vote 30 down vote accepted

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

$$\sum_{i=1}^\infty x_i$$


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 $f \to \infty$.

share|cite|improve this answer
$\dfrac{a_{n+1}-a_n}{a_n} = \dfrac{\Delta a_n}{a_n}$, so it does look like $\dfrac{df}{f}$, but if I had answered this the answer would not have been so efficient. – Michael Hardy Nov 13 '13 at 18:54
@MichaelHardy What do you mean? Thanks for the edit btw. You were right about the French! – Bruno Joyal Nov 13 '13 at 18:55
wow. such nice. – Aryabhata Nov 13 '13 at 19:36
the integral has to be from $ 1$ to infinity – toufik_kh.17 Apr 11 '14 at 5:53
@toufik_kh.17 The analogy here is that the forward difference operator on sequences is analogous to the derivative. It's just an analogy, don't try to read too much into it... – Bruno Joyal Apr 11 '14 at 22:46

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.