Determine whether the following sequence is increasing, decreasing, or not monotonic. Is the sequence bounded?
(a) a_n = cos(n*pi / 2) (b) a_n = n/(n^2 + 1)
first one i would just put in n = 1, 2, 3, and see what you get
it is certainly bounded because cosine is bounded above by 1 and below by -1 did you try plugging in the numbers? do 4 and it will be very clear
Oh I see, but what is the part b bounded by?
I know it decreases but I still need to find the bounds.
did you get 0,-1,0,1,0,-1,... for the first one?
For the monotonic sequence theorem, does the sequence have to be bounded on BOTH sides? (top and bottom)?
it is bounded below by 0 for sure
wait are you talking about question 1 or 2?
and yes I got that.... for part a. I got the same pattern.
ok so on to part b
What is part b bounded by? Yes ok
you know the limit is zero right?
yes
That is only the lower bound. Don't we need to find an upper bound also?
are you trying to show it is monotone and bounded?
Yes^
ok so it is bounded above by \(\frac{1}{2}\) but that is not important it is bounded below by 0 because the terms are all positive so what is left is to see if it is monotone decreasing
and it does decrease. how did you receive the bounded above part?
also does the sequence have to be bounded ABOVE and BELOW to be counted as bounded?
i replaced n by 1 if it is decreasing it is largest when n is smallest
n could be negative numbers though, correct?
no
????
\(a_n\) n is a positive integer
\[a_1,a_2,a_3,a_4,...\]
otherwise nothing would be anything if you could go to either plus or minus infinity, what a mess that would be
Sounds good, thanks
the easiest way to show the sequence is monotone decreasing is the take the derivative of the corresponding function \[f(x)=\frac{x}{x^2+1}\] and show that if \(x>1\) the derivative is negative, so the function is decreasing and therefore so is the sequence
otherwise it is very tedious to show it is decreasing, even though it is obvious
I getcha, thanks
yw
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