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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!

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marked as duplicate by Did sequences-and-series May 2 '16 at 22:18

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    $\begingroup$ ...and dudettes! $\endgroup$ – Bruno Joyal Nov 13 '13 at 18:49
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    $\begingroup$ @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)? $\endgroup$ – Sudhir Nov 14 '13 at 16:46
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    $\begingroup$ Dear Sudhir: in the question, the $a_n$'s are integers. $\endgroup$ – Bruno Joyal Nov 14 '13 at 18:27
  • $\begingroup$ Consider a partial sum: $\displaystyle{\large S_{N} \equiv \sum_{n = 1}^{N}{a_{n + 1} - a_{n} \over a_{n}} = \sum_{n = 1}^{N}{a_{n + 1} \over a_{n}} - N > 0}$. $\endgroup$ – Felix Marin Nov 15 '13 at 4:23
  • $\begingroup$ @FelixMarin And so what? Your comment seems like a misleading hint to me. If it is not, please explain. $\endgroup$ – Did May 5 '16 at 9:19
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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 $f \to \infty$.

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  • $\begingroup$ $\dfrac{a_{n+1}-a_n}{a_n} = \dfrac{\Delta a_n}{a_n}$, so it does look like $\dfrac{df}{f}$, but if I had answered this the answer would not have been so efficient. $\endgroup$ – Michael Hardy Nov 13 '13 at 18:54
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    $\begingroup$ @MichaelHardy What do you mean? Thanks for the edit btw. You were right about the French! $\endgroup$ – Bruno Joyal Nov 13 '13 at 18:55
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    $\begingroup$ wow. such nice. $\endgroup$ – Aryabhata Nov 13 '13 at 19:36
  • $\begingroup$ the integral has to be from $ 1$ to infinity $\endgroup$ – toufik_kh.17 Apr 11 '14 at 5:53
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    $\begingroup$ @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... $\endgroup$ – Bruno Joyal Apr 11 '14 at 22:46
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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.

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  • $\begingroup$ The key is that there is a constant $c > 0$ such that $a_{n+1} \ge a_n+c$. $\endgroup$ – marty cohen May 2 '16 at 22:19

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