A proof of a property of limits

Question

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?

Answer 1

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 $n\to \infty$

Answer 2

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 $.

Answer 3

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.

Answer 4

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.)

Answer 5

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.

Answer 6

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

Answer 7

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 $b_n = a_n +1$.