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?
On
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 $b_n = a_n +1$.
On
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 $.
On
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 $\lim_{n\to\infty}d_n=\pm\infty$.
On
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.)
On
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.
On
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.
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 $n\to \infty$