Question

Consider the sequence {an}={3n+13n−3n3n+1}. Graph this sequence and use your graph to help you answer the following questions.

Answer 1

Part 1: You can simplify $a_n$ to

$\dfrac{3n+1}{3n}-\dfrac{3n}{3n+1} = \dfrac1{3n}+\dfrac1{3n+1}$

Presumably, the sequence starts at n = 1. It's easy to see that the sequence is strictly decreasing, since larger values of n make either fraction smaller.

(a) So, the sequence is bounded above by its first value,

$|a_n| \le a_1 = \dfrac13+\dfrac14 = \boxed{\dfrac7{12}}$

(b) And because both fractions in $a_n$ converge to 0, while remaining positive for any natural number n, the sequence is bounded below by 0,

$|a_n| \ge \boxed{0}$

(c) Finally, $a_n$ is bounded above and below, so it is a bounded sequence.

Part 2: Yes, $a_n$ is monotonic and strictly decreasing.

Part 3:

(a) I assume the choices are between convergent and divergent. Any monotonic and bounded sequence is convergent.

(b) Since $a_n$ is decreasing and bounded below by 0, its limit as n goes to infinity is 0.

Part 4:

(a) We have

$\displaystyle \lim_{n\to\infty} \frac{10n^2+1}{n^2+n} = \lim_{n\to\infty}10+\frac1{n^2}}{1+\frac1n} = 10$

and the (-1)ⁿ makes this limit alternate between -10 and 10. So the sequence is bounded but clearly not monotonic, and hence divergent.

(b) Taking the limit gives

$\displaystyle\lim_{n\to\infty}\frac{10n^3+1}{n^2+n} = \lim_{n\to\infty}\frac{10+\frac1{n^3}}{\frac1n+\frac1{n^2}} = \infty$

so the sequence is unbounded and divergent. It should also be easy to see or establish that the sequence is strictly increasing and thus monotonic.

For the next three, I'm guessing the options here are something to the effect of "does", "may", or "does not".

(c) may : the sequence in (a) demonstrates that a bounded sequence need not converge

(d) does not : a monotonic sequence has to be bounded in order to converge, otherwise it grows to ± infinity.

(e) does : this is true and is known as the monotone convergence theorem.