The sequence $\left\{ \frac{5}{n} \right\}_{n=1}^\infty$ is not monotonic, or not convergent, or bounded. Why?

Question

I recently watched a video where a Calculus instructor asked the viewers to try to come up with a monotonic, unbounded, convergent sequence, if they could, as an exercise to understand these concepts better.

I came up with $$\{a_n\}_{n=1}^\infty = \left\{ \frac{5}{n} \right\}_{n=1}^\infty$$

1. I claim it is monotonic because because the sequence is decreasing.

2. I claim it is convergent because as $n \to \infty, \, a_n \to 0$.

3. I claim it is unbounded because for every one of its members, we can always find a smaller one.

However, right after the exercise, we were presented with the following theorem:

If a sequence is convergent, then it is bounded.

Hence my solution must be wrong, and at least one of my claims must be false. But I'm not sure which one, or why. Is anyone able to shed some light on this?

Answer 1

You are correct that the sequence is monotonically decreasing.

It is convergent, as you say.

An upper bound is just a number that is at least as large as all the members of the sequence, and a lower bound is just a number at least as small as all the members of the sequence. $1000$ is an upper bound and $-1000$ is a lower bound, so the sequence is bounded. There are tighter bounds, $5$ and $0$, but we were not asked for the greatest lower bound and least upper bound.

Answer 2

It's bounded by 0 below and 5 above.

Answer 3

The sequence is monotonically decreasing, converges to $0$, and is bounded above by $5$ and below by $0$.

Answer 4

Your 3rd claim is wrong as all above have suggested. Also, the reasoning would follow this order:

• the sequence is monotonically decreasing

• the sequence is bounded above by $5$ and below by $0$

• hence by monotone-bounded theorem,the sequence converges to a real number,in this case to $0$.