# Math Help - Convergence problem

1. ## Convergence problem

This is a problem on convergence that was confusing me, so any help would be appreciated!

*Note that when I say Sigma below, I mean the sigma symbol (I'm not sure how to input it so I just wrote the word)

Suppose Sigma starting with j = 0 and ending at infinity of b_j converges and a_j > b_j for all j>16. What can we conclude about Sigma starting with j = 0 and ending at infinity for a_j?

(Meaning, does it converge, diverge, etc?)

2. Originally Posted by clockingly
This is a problem on convergence that was confusing me, so any help would be appreciated!

*Note that when I say Sigma below, I mean the sigma symbol (I'm not sure how to input it so I just wrote the word)

Suppose Sigma starting with j = 0 and ending at infinity converges and a_j > b_j for all j>16. What can we conclude about Sigma starting with j = 0 and ending at infinity for a_j?

(Meaning, does it converge, diverge, etc?)
please tell us what series converges. Sigma starting with j = 0 and ending at infinity (OF WHAT) converges

3. Originally Posted by Jhevon
please tell us what series converges. Sigma starting with j = 0 and ending at infinity (OF WHAT) converges
Ah, sorry - I meant to write it, but didn't by accident (I have corrected it in my first post).

4. Originally Posted by clockingly
Ah, sorry - I meant to write it, but didn't by accident (I have corrected it in my first post).
Well, we can't say anything about the infinite series a_j on that basis. If b_j DIVERGED, then we could be sure that a_j diverged by the comparison test. However, if b_j converges, a_j can either converge or diverge if a_j > b_j.

Now if b_j > a_j, then we'd be in business

5. Originally Posted by Jhevon
Well, we can't say anything about the infinite series a_j on that basis. If b_j DIVERGED, then we could be sure that a_j diverged by the comparison test. However, if b_j converges, a_j can either converge or diverge if a_j > b_j.

Now if b_j > a_j, then we'd be in business
Divergent is not sufficient.
You need the fact that is is divergent to +oo or -oo.

6. Originally Posted by ThePerfectHacker
Divergent is not sufficient.
You need the fact that is is divergent to +oo or -oo.
i was under the impression that "divergent" means it goes to +oo or -oo

7. Originally Posted by Jhevon
i was under the impression that "divergent" means it goes to +oo or -oo
But divergent means not convergent.

8. Originally Posted by ThePerfectHacker
But divergent means not convergent.
Yes, i agree.

To my knowledge there are three possibilities for an infinite series:
1) it goes to some real number
2) it goes to +oo
3) it goes to -oo

in case 1) we say it converges, in cases 2) and 3) we say it diverges. am i mistaken somewhere?

9. Originally Posted by Jhevon
Yes, i agree.

To my knowledge there are three possibilities for an infinite series:
1) it goes to some real number
2) it goes to +oo
3) it goes to -oo

in case 1) we say it converges, in cases 2) and 3) we say it diverges. am i mistaken somewhere?

SUM(n=1,+oo) (-1)^n?

10. Originally Posted by ThePerfectHacker

SUM(n=1,+oo) (-1)^n?
it converges to some real number, though at the moment, i'm not sure what

i'm tempted to say 0, but something in my gut says it won't be so simple

11. Originally Posted by Jhevon
it converges to some real number, though at the moment, i'm not sure what

i'm tempted to say 0, but something in my gut says it won't be so simple
Remember the definition what what is means convergence of a series (we had it on our mid-term).

Consider the sequence of partial sums,

S_1=-1
S_2=-1+1=0
S_3=-1+1-1=-1
And so on,

Thus,
-1,0,-1,0,-1,...
Is the sequence of partial sums.

This sequence has no limit.

There are two ways to show it has no limit.
Either violate the epsilon definition.

But here is an easier way.
Assume that is has a limit (real or infinite).
Then, all subsequences must have the same limit (theorem 11.2).

But,
a_even = 0,0,0,0,...
b_odd = -1,-1,-1,...
Are two subsequences that converge to two different values, i.e. -1 and 0 irrespectively.

Thus, there is no limit.

12. Originally Posted by ThePerfectHacker
Remember the definition what what is means convergence of a series (we had it on our mid-term).

Consider the sequence of partial sums,

S_1=-1
S_2=-1+1=0
S_3=-1+1-1=-1
And so on,

Thus,
-1,0,-1,0,-1,...
Is the sequence of partial sums.

This sequence has no limit.

There are two ways to show it has no limit.
Either violate the epsilon definition.

But here is an easier way.
Assume that is has a limit (real or infinite).
Then, all subsequences must have the same limit (theorem 11.2).

But,
a_even = 0,0,0,0,...
b_odd = -1,-1,-1,...
Are two subsequences that converge to two different values, i.e. -1 and 0 irrespectively.

Thus, there is no limit.
Ah yes, ok.

somehow i mixed up sequences and series when i was thinking about the problem.