hy23 said:
wait, in your case how is it that 100 is a peak when 899 is greater than 100? Or do you mean that 100 only has to be bigger than the one subsequent term (99)?
Great! You caught me napping

.
I made a mistake sorry for the confusion.
the other thing is I get what you are saying about how after a while, the sequence either flattens out and I can see that clearly it will converge, but you also say that if it doesn't stay constant it increases and if it has an upper bound then it converges, this is what I don't understand because how do you know that there won't be more peaks, that after it increases it might decrease in an oscillatory fashion
What I was saying is that if it has
finitely many peaks then it is either evetually constant or eventually increasing. The word
finite is very important because sequences are
infinite in nature.
If the sequence decreases later in an oscilatory fashion then you have peaks! If the oscillatory decrease continues "forever" then you have
infinitly many peaks. If it stops after a while, it means that ( i.e assume the oscillation occurs from n =1000 to 50,000 , after 50,000 the sequence is well behaved) after a while the sequence behaves itself.
Remember in the study of sequences the behaviour as n goes to
infinity is what matters, not what happens at between say 1,000,000 and 24,000,000.
do you have any good examples of a sequence that is not a trig function that has a convergent subsequence, an example that really demonstrates the proof nicely might put my brain to rest
thanks!
Trig are nice examples because of the oscillation but consider the sequence...
s_n = \frac{1}{n} \quad for \quad odd \quad n \quad divisible \quad by \quad 5
s_n = -4 + \frac{1}{\sqrt{n}} \quad for \quad odd \quad n \quad not \quad divisible \quad by \quad 5
s_n= -e^{ \frac{1}{n} } \quad for \quad even \quad n
This sequence has 3 convergence subsequences but there are weird twist and turns everywhere.