MHB Can a sequence without a convergent subsequence have a limit of infinity?

AI Thread Summary
A sequence of positive real numbers without a convergent subsequence must be unbounded, leading to the conclusion that its limit as n approaches infinity is positive infinity. However, if the sequence includes both positive and negative values, the absolute values of the terms must also approach infinity. The discussion highlights that being unbounded does not guarantee that the limit is infinity, as demonstrated by the sequence x_n = |tan(n)|, which oscillates and does not converge. The clarification emphasizes that for a sequence to have a limit of infinity, it must not have a bounded subsequence. Thus, the relationship between boundedness and convergence is crucial in determining the limit behavior of sequences.
Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
I quote a question from Yahoo! Answers

Let (xn)n be a sequence of positive real numbers that has no convergent subsequence. Prove that lim(n→∞) x of n=+∞. What if the xn are permitted to take both positive and negative values?

I have given a link to the topic there so the OP can see my response.
 
Mathematics news on Phys.org
Each bounded sequence has a convergent subsequence, so $x_n$ has no convergent subsequence implies that $x_n$ is not bounded. In this case, $x_n\to +\infty$ if $x_n$ is a sequence of positive real numbers and $\left|x_n\right|\to +\infty$ if the $x_n$ are permitted to take both positive and negative values.
 
Fernando Revilla said:
Each bounded sequence has a convergent subsequence, so $x_n$ has no convergent subsequence implies that $x_n$ is not bounded. In this case, $x_n\to +\infty$ if $x_n$ is a sequence of positive real numbers and $\left|x_n\right|\to +\infty$...

The fact that the positive term sequence $x_{n}$ is unbounded doesn't mean that $\displaystyle \lim_{n \rightarrow \infty} x_{n} = + \infty$. As example You can consider the sequence $\displaystyle x_{n} = |\tan n|,\ n \ge 1$...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
The fact that the positive term sequence $x_{n}$ is unbounded doesn't mean that $\displaystyle \lim_{n \rightarrow \infty} x_{n} = + \infty$. As example You can consider the sequence $\displaystyle x_{n} = |\tan n|,\ n \ge 1$...

Right, my fault, I was thinking about a monotic sequence.
 
The negation of $\lim_{n\to\infty}x_n=\infty$ says that $x_n$ has a bounded subsequence and therefore a convergent sub-subsequence.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top