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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?

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What do you mean by the limit of $\frac{d_{n+1}}{d_{n}}$ when $x$ approaches $\infty$? Should it be $n\to\infty$? –  Thomas E. 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
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$d_n = 1 - 1/n$ is a counterexample... –  Adam Rubinson Jan 10 '13 at 20:23
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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

7 Answers 7

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 $n\to \infty$

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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 $.

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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.

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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.)

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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.

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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 $\lim_{n\to\infty}d_n=\pm\infty$.

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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 $b_n = a_n +1$.

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

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