# Convergent and Divergent Sequences

1. Jan 24, 2014

### 939

1. The problem statement, all variables and given/known data

Please look over my work and tell me if I did something wrong.

Suppose Bn is a divergent sequence with the limit +∞, and c is a constant.

Prove: lim cBn -> ∞ = +∞ for c > 0

2. Relevant equations

N/A

3. The attempt at a solution

lim Bn -> ∞ = means that for some value K > 0, Bn > K for all n > N.

If c > 0, multiplying all terms by a constant c > 0 will not change the fact of a limit of +∞.

cK > 0, cBn > cK for all n > N meets the definition of a divergent sequence with a limit of +∞ and thus, if Bn is a divergent sequence with the limit +∞, and c is a positive constant:

lim cBn -> ∞ = +∞

....

With an example (c = 1/2)
lim n -> ∞ = n/2
= lim n -> ∞ (1/2)(n) = +∞

2. Jan 24, 2014

### Dick

You should start with the correct definition of the limit. lim Bn -> ∞ = means that FOR ALL values K > 0, THERE EXISTS an N such that Bn > K for all n > N. Use that to show that there is a possibly different N that works for cBn.

3. Jan 24, 2014

### 939

Thanks for the help!

Would it suffice to show that cN and cn work?

4. Jan 24, 2014

### LCKurtz

What do you mean by cN and cn "work"? Given $K > 0$ you have to show how to find an $N_1$ such that if $n>N_1$ then $cb_n > K$.

Last edited: Jan 24, 2014
5. Jan 25, 2014

### 939

Thanks.

My final question is if you could tell me what exactly K, n, N, and bn represent on the graph (i.e. horizontal or vertical) just to make sure I'm not mistaken.

6. Jan 25, 2014

### Dick

Mistaken about what? You really still haven't really addressed the problem.

7. Jan 26, 2014

### 939

Thanks, but I'm trying to picture what exactly $n>N_1$, for example, or $b_n > K$ mean on a graph so it's easier to picture.

8. Jan 26, 2014

### Dick

There's not really much to picture. Look at http://en.wikipedia.org/wiki/Limit_of_a_sequence. They have a graph. To graph $b_n$ you just put a dot at $y=b_n$ and $x=n$ for all n for which the sequence is defined.

Maybe better to think of an example. Suppose $b_n=n^2$ and they give you $K=10000$. You should be able to figure out that $N=100$ is a good value for N since $b_n=n^2>K=10000$ if $n>N=100$. Now take c=1/100. So $c_n=n^2/100$. What's an N corresponding to K=10000 for that series?

9. Jan 26, 2014

### 939

Thanks! I believe N = 1000 works for the series $c_n=n^2/100$.

10. Jan 26, 2014

### Dick

Sure it does. Now forget about the n^2 example where you can just calculate things. Go back to the original question. If you want $c b_n>K$ then how large should $b_n$ be?

11. Jan 26, 2014

### 939

Bn should be a value > (k/c)?

12. Jan 26, 2014

### Dick

Yes. Now since Bn->infinity you know there is a value of N such that Bn>(K/c) for all n>N. What does that tell you about the divergence of cBn?

Last edited: Jan 26, 2014
13. Jan 26, 2014

### 939

It tells me that cBn is a divergent sequence with the limit +∞, just like bn was, and thus it is true?

14. Jan 26, 2014

### Dick

Yes, now write the whole thing out in the form of a proof. To prove lim cBn=+∞, you want to show given any K, there is an N such that cBn>K for all n>N. Fill in the argument as to why.

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