Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Today during lecture my lecturer showed us this property, but provided no proof.

If $$\lim_{n\to\infty} {d_{n+1}\over d_n} >1$$ then $$\lim_{n\to\infty}d_{n}=\infty $$

Is this property legit? (not to be disrespectful to my lecturer but he tends to make a lot of mistakes)

And if it is, what is the logic behind that property? How does it behave when the first limit tends to 1 or is less than 1?

share|cite|improve this question
What do you mean by the limit of $\frac{d_{n+1}}{d_{n}}$ when $x$ approaches $\infty$? Should it be $n\to\infty$? – T. Eskin Jan 10 '13 at 19:15
Yes, you're right. It's $$n\to\infty$$ – TerryGoldenstein Jan 10 '13 at 19:19
Think about a sequence where the next number is about twice the previous one. Where does it go? What if every time the numbers are cut by 1/2? – Maesumi Jan 10 '13 at 19:26
$d_n = 1 - 1/n$ is a counterexample... – Adam Rubinson Jan 10 '13 at 20:23
No hold on, $d_{n+1} / d_n$ tends to 1 (which is not greater than 1) in my example. My bad. – Adam Rubinson Jan 10 '13 at 20:55
up vote 8 down vote accepted

Assume that this is a positive sequence. (You might have $\lim_{n\to \infty} d_n = -\infty$). There is a $M$ and $\delta > 0$ such that for $n\geq M$ $$ d_{n+1}/d_n > 1 + \delta = a > 1. $$ That is: $$ d_{n+1} > ad_n. $$ So for $n> M$: $$d_n > ad_{n-1} > a^2d_{n-2}... > a^{n-M}d_M.$$ Now let $n\to \infty$

share|cite|improve this answer
Thanks, very nice reasoning. I might sound naive but how does one develop such familiarity with proving theorems/properties? – TerryGoldenstein Jan 10 '13 at 19:46
@TerryGoldenstein: I am sure that there are many theories out there about how people learn, but I think that it really all comes down to practice - practice - practice. That said, as with many things, it is good to try a lot on your own. Over time you start to see patterns showing up. – Thomas Jan 10 '13 at 19:54

Since we have $\lim_{n \to \infty } d_{n+1}/d_n > 1$,

Let us have $ \delta = \min \{ d_{n+1} - d_n: n\ge N \text{ for some N }\in \mathbb N\}$, then we have $ \lim_{n \to \infty} d_n > \lim_{n \to \infty} d_N + n\delta$ which diverges to $\infty $.

share|cite|improve this answer
can you help me understand why $\lim_{n\to\infty} d_n > \lim_{n\to\infty} d_N + n\delta$? – Rustyn Jan 10 '13 at 20:49
@RustynYazdanpour since $\delta $ is the minimum of difference, we have $<$ and we know that $d_{n+1} = d_n + \delta $, but there are infinite terms ... so we have $n \delta $ – Santosh Linkha Jan 10 '13 at 20:53
that is helpful thx – Rustyn Jan 10 '13 at 20:55
you are welcome :) – Santosh Linkha Jan 10 '13 at 20:56

If $\lim_{n\to\infty}\frac{d_{n+1}}{d_n}>1$ then for there is a $\epsilon>0$ and $N\in\mathbb{N}$ such that $n>N$ implies $\frac{d_{n+1}}{d_n}>1+\epsilon\implies e^{d_{n+1}-d_{n}}>e^{0+\epsilon^\prime}\implies |d_{n+1}-d_n|>\epsilon^\prime $. Then $$ \lim_{n\to\infty}|d_{n+1}-d_n|>\epsilon^\prime $$ But this contradicts the criterion of convergence of sequences cauchy.

share|cite|improve this answer

If the sequence converges $d_n\to L$, then eventually its terms must be almost all the same, so their ratios should approach $1$. (I'm glossing over what happens if $L = 0,$ by the way -- this is just intuition.)

share|cite|improve this answer

Since the limit of ratios is greater than 1, eventually all terms have the same sign. If the sign is negative, then $d_n \to - \infty.$ For example consider $d_n=-(2^n)$, then $d_{n+1}/d_n=2,$ a constant, making the limit 2 as required, yet the terms approach $- \infty$ instead of $\infty$.

If the terms are eventually positive the conclusion follows, since if the limit is $a>1$ we can choose $c>1$ with also $c<a$ and eventually for $n \ge N$ we will have

[1] $d_N>0$

[2] $d_{N+k}>c^k\ d_N$

(where [2] is shown by induction). Then since $c^k \to \infty$ we have that $d_n$ approaches infinity.

share|cite|improve this answer

If $\lim_{n\to\infty}d_n=L$ where $L\in\mathbb{R}$, then $\lim_{n\to\infty}\frac{d_{n+1}}{d_n}=1$. (Can you see why?)

Thus if $\lim_{n\to\infty}\frac{d_{n+1}}{d_n}\ne1$ we know $(d_n)_{n=1}^{\infty}$ diverges.

If $\lim_{n\to\infty}\frac{d_{n+1}}{d_n}>1$ then, for the most part $|d_{n+1}|>|d_n|$, so $\lim_{n\to\infty}d_n=\pm\infty$.

share|cite|improve this answer
If for all $n$ $d_n = 0$, then what is $\lim_{n\to\infty} \frac{d_{n+1}}{d_n}$? – Neal Jan 10 '13 at 19:37

Assuming your hypotheses: $$ \lim_{n\to \infty} \dfrac{d_{n+1}}{d_n} > 1 $$ Now later assume that $\lim_{n\to\infty} d_n = L$.
We assume here that $L\ne 0$ and that $d_{n} \ne 0$, $\forall n \ge m$. Recall from (Limit Laws): $$ \lim_{n\to\infty} \dfrac{d_{n+1}}{d_n} = \dfrac{\lim_{n\to\infty}d_n}{\lim_{n\to\infty}d_{n+1}} = \dfrac{L}{L} = 1 $$ which is a contradiction.
If $L = 0$, repeat same argument with $b_n = a_n +1$.

share|cite|improve this answer
This isn't enough to conclude the sequence diverges; monotonic increasing sequences converge as long as they are bounded. You need to show unboundedness. – Gyu Eun Lee Jan 10 '13 at 19:31
You argument does not answer why the sequence could not converge to some limit. – Tomás Jan 10 '13 at 19:32

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.