1. ## Sequence Convergence

I have two sequences:

$\displaystyle {0,1,0,0,1,0,0,0,1...}$

$\displaystyle {\frac{1}{1},\frac{1}{3},\frac{1}{2},\frac{1}{4},\ frac{1}{3},\frac{1}{5},\frac{1}{4},\frac{1}{6},... }$

By the Monotonic Sequence Theorem, I think that these are both convergent. The first sequence is bounded between 1 and 0 and is monotonic. Therefore it must converge right? But where? At 0?

The same seems to be true with the second sequence. The second sequence is also bounded between 1 and 0. It doesn't fit the definition of decreasing or increasing since it jumps up and down, so it must be monotonic. Yet it's gradually approaching a value of zero.

I have two sequences:

$\displaystyle {0,1,0,0,1,0,0,0,1...}$

$\displaystyle {\frac{1}{1},\frac{1}{3},\frac{1}{2},\frac{1}{4},\ frac{1}{3},\frac{1}{5},\frac{1}{4},\frac{1}{6},... }$

By the Monotonic Sequence Theorem, I think that these are both convergent. The first sequence is bounded between 1 and 0 and is monotonic. Therefore it must converge right? But where? At 0?

The same seems to be true with the second sequence. The second sequence is also bounded between 1 and 0. It doesn't fit the definition of decreasing or increasing since it jumps up and down, so it must be monotonic. Yet it's gradually approaching a value of zero.
What makes you think that the first is monotonic?

3. Originally Posted by Drexel28
What makes you think that the first is monotonic?
Ok, I actually misread the definition of monotonic. A sequence is monotonic if it is either increasing or decreasing. I accidently read that as Neither

So it's not monotonic. It doesn't have a limit, so it is divergent correct?

But for the second sequence, it doesn't seem to fit the definition of monotonic either. Even though it is decreasing in general, it jumps up and down. So I can't say that $\displaystyle a_n>a_{n+1},\forall n$.

But for the second sequence, it doesn't seem to fit the definition of monotonic either. Even though it is decreasing in general, it jumps up and down. So I can't say that $\displaystyle a_n>a_{n+1},\forall n$.