Showing a sequence converges or diverges.

Ethan3141

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Hey, I was just wondering if anyone knows the name of this theorem/proposition (not exact wording since I don't know where to find it):

if A<=B and B converges then A converges,
if A>=B and B diverges then A diverges
(A and B are sequences)

It looks similar to the squeeze theorem I guess but other than that I don't know what it is called. Any help would be appreciated.

Thanks, Ethan.
 
Anyone knows the name of this theorem/proposition (not exact wording since I don't know where to find it):
if A<=B and B converges then A converges,
if A>=B and B diverges then A diverges
(A and B are sequences)
It looks similar to the squeeze theorem I guess but other than that I don't know what it is called. Any help would be appreciated.
You notation is very poor. Here is a guess as to what A<=B means.
Suppose that \(\displaystyle a_n=(-1)^n+\frac{1}{n}\), here are a few terms of the A sequence: \(\displaystyle 0,~\frac{3}{2},~\frac{-2}{3},~\frac{5}{4},~\frac{-4}{5}\cdots\)
If the B sequence is \(\displaystyle b_n=5+\frac{1}{n}\) the \(\displaystyle (\forall n)[~ a_n\le b_n~]\)
The B sequence converges while the A sequence does not converge. Both are bounded sequences
For the second question just swap the letters and use the same idea.
 
You notation is very poor. Here is a guess as to what A<=B means.
Suppose that \(\displaystyle a_n=(-1)^n+\frac{1}{n}\), here are a few terms of the A sequence: \(\displaystyle 0,~\frac{3}{2},~\frac{-2}{3},~\frac{5}{4},~\frac{-4}{5}\cdots\)
If the B sequence is \(\displaystyle b_n=5+\frac{1}{n}\) the \(\displaystyle (\forall n)[~ a_n\le b_n~]\)
The B sequence converges while the A sequence does not converge. Both are bounded sequences
For the second question just swap the letters and use the same idea.

Yeah you're right my notation is poor, <= (every term of A is less than or equal to their counterpart in B) means less than or equal to and I am certain that this fact held true when I used it, this being said I suppose it must have the restraint that both sequences must be positive. It is similar to the direct comparison test but that is for series' only. Sorry for the poor notation and thanks for the counter example.
 
Yeah you're right my notation is poor, <= (every term of A is less than or equal to their counterpart in B) means less than or equal to and I am certain that this fact held true when I used it, this being said I suppose it must have the restraint that both sequences must be positive. It is similar to the direct comparison test but that is for series' only. Sorry for the poor notation and thanks for the counter example.
\(\displaystyle a_n =1+\frac{1}{n} ~~ \text{n is odd} \\a_n= 2+\frac{1}{n} ~~ \text{n is even}\)
Both are positive and \(\displaystyle a_n\le b_n\)
 
\(\displaystyle a_n =1+\frac{1}{n} ~~ \text{n is odd} \\a_n= 2+\frac{1}{n} ~~ \text{n is even}\)
Both are positive and \(\displaystyle a_n\le b_n\)
Ok, then I guess it has to be for all n in the natural number set that the inequality holds. Starting to doubt this even holds anymore, lol. Especially considering I used it to show that the sequence x_n = sqrt(x) - sqrt(x+1) converges (which obviously has negative terms).

And your grammar is very poor indeed.
If this is aimed at me I agree, I really should get some software to help with it since I am dyslexic but I have never bothered and probably never will (I feel like it takes the authenticity away from what I type). Since it wasn't aimed at me I guess this is now a confession that I have bad grammar.
 
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Ok, then I guess it has to be for all n in the natural number set that the inequality holds. Starting to doubt this even holds anymore, lol.


If this is aimed at me I agree, I really should get some software to help with it since I am dyslexic but I have never bothered and probably never will (I feel like it takes the authenticity away from what I type).
No, no, no. That was for @pka.
 
Ok, then I guess it has to be for all n in the natural number set that the inequality holds. Starting to doubt this even holds anymore, lol.
That is the whole point. it does not hold.
There is a vast difference in properties of sequences and series.
Series are sequences of initial partial sums of a sequence.
Consider the sequence \(\displaystyle a_n\) then define \(\displaystyle S_N= \sum\limits_{n = 1}^N {{a_n}} \)
Now we have a series that is a sequence of initial partial sums. If the \(\displaystyle a_n\cancel{\to}0 \)(i.e. is not null) then the series must diverge.
However, if the sequence \(\displaystyle a_n\to 0\) is no assurance that the series does converge.
 
@pka, you have a math processing error in line 5 of your last post.
No there is no error that only you limited knowledge of series or perhaps notation .
That is often called the first test for convergence. If the associated sequence of a series, \(\displaystyle \sum\limits_{n = 1}^\infty {{x_n}}\), is not null (i.e, \(\displaystyle x_n\cancel{\to}0)\) then the series diverges.
 
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