Convergent and Divergent Sequences

[939]

Homework Statement

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

Homework Equations

N/A

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) = +∞

 

[Dick]

Homework Statement

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

Homework Equations

N/A

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) = +∞

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.

 

[939]

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.

Thanks for the help!

Would it suffice to show that cN and cn work?

 

[LCKurtz]

Thanks for the help!

Would it suffice to show that cN and cn work?

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 $cb_n > K$.

 

[939]

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 $cb_n > K$.

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.

 

[Dick]

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.

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

 

[939]

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

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.

 

[Dick]

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.

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?

 

[939]

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?

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

 

[Dick]

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

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?

 

[939]

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?

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

 

[Dick]

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

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?

 

[939]

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?

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

 

[Dick]

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

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.