If $(x_n)$ is a sequence of positive real nunbers such that $\lim_{x \to \infty} \frac{x_{n+1}}{x_n} = L > 1$ how to show $x_n$ diverges?
#Sequence test
nocturne lion
tulip spadeBOT
Halex
white summitBOT
If $(x_n)$ is a sequence of positive real nunbers such that $\lim_{x \to \infty}...
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nocturne lion
Assume $x_n$ converges to $a$, then $$x_{n+1} = \frac{x_{n+1}}{x_n} * x_n$$ and taking limit both sides we get $$a = La \implies L=1$$ which is a contradiction
tulip spadeBOT
Halex
nocturne lion
Is this right
tender topaz
If $(x_n)$ is a sequence of positive real nunbers such that $\lim_{x \to \infty}...
If that limit is L, then x(n) behaves like L^n for large n. And since L > 1, what can we say about the behavior of L^n?
nocturne lion
If that limit is L, then x(n) behaves like L^n for large n. And since L > 1, wha...
How comes x_n behaves like L^n?
tender topaz
How comes x_n behaves like L^n?
a(n) = L^n (perhaps, multiplied by a constant, but that doesn't matter) is the sequence that exactly satisfied a(n + 1)/a(n) = L.
Here it's in the limit, so it's not equal to that, but it still behaves like that.
nocturne lion
Here it's in the limit, so it's not equal to that, but it still behaves like tha...
Oh yeah that's an example but still stuck at proving this
agile locust
Is this right
provided a is positive
but a=0 is also a possibility
@nocturne lion
If we assume for a contradiction x_n converges to 0 then for all k > N we have x_{n+1}/x_n > 1
where N is large enough
then
$$\frac{|x_k|}{|x_N|} = \frac{|x_k|}{|x_{k-1}|}\ldots\frac{|x_{N+1}|}{|x_N|} > 1$$
tulip spadeBOT
aL
agile locust
but then $$|x_k| > |x_N| > 0$$
tulip spadeBOT
aL
agile locust
contradicting the assumption that x_k converges to 0
this is the same argument we can use to prove the d'Alembert test for divergence of series
violet wadi
Set $k=\frac{L+1}2>1.$ There exists $N$ such that $x_{n+1}>k\ x_n$ for $n\ge N$, hence $x_{n+m}>k^m\ x_n.$ As $k^m$ can be made arbitrary large, $(x_n)$ tends to $+\cal1$.
tulip spadeBOT
mimshell
nocturne lion
If we assume for a contradiction x_n converges to 0 then for all k > N we have x...
are you working in the other case?
nocturne lion
$$\frac{|x_k|}{|x_N|} = \frac{|x_k|}{|x_{k-1}|}\ldots\frac{|x_{N+1}|}{|x_N|} > 1...
why is this?
nocturne lion
Set $k=\frac{L+1}2>1.$ There exists $N$ such that $x_{n+1}>k\ x_n$ for $n\ge N$,...
I tried a similar approach:
For $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $$\forall n> N \implies \left| \frac{x_{n+1}}{x_n} - L < \epsilon \right| , $$ taking $\epsilon = \frac{L-1}{2} > 0$, we can deduce that $$\frac{L+1}{2} < \frac{x_{n+1}}{x_n}$$
agile locust
why is this?
because by assumption the ratio converges to L > 1, so eventually the ratios must be larger than 1+\delta for some delta > 0
tulip spadeBOT
Halex
nocturne lion
I tried a similar approach:
For $\epsilon > 0$, there exists $N \in \mathbb{N}$ ...
We can show by induction that $$\left( \frac{L+1}{2} \right)^n < x_n$$ for $n$ arbitrary large, and since $\frac{L+1}{2} > 1$ this sequence on the left diverges, proving that $x_n$ diverges as well. Not sure if I did somethig work
tulip spadeBOT
Halex
nocturne lion
n > N
nocturne lion
how do you go from this
I used induction to prove it works
nocturne lion
@agile locust so, do you consider that what I did is correct?
abstract havenBOT
@nocturne lion
abstract havenBOT
@agile locust @violet wadi @tender topaz The user still needs help with this help request.
nocturne lion
+close
abstract havenBOT
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