Testing a sequence for converge or divergence

[shemer77]

I have 4 problems left and the questions says I have to test them for converge or divergence.
Here are the problems
http://gyazo.com/f0fa5a38c5968ecb7e74103486a181bd.png
http://gyazo.com/4689f9d02d0b264c2c2b64ff4907ba77.png
for 25, I want to take the limit as n goes to infinity however I get (-1)^\infty and stuck there

for 27, similar problem how do i take the limit of cos(\infty)

for 43 i have no idea

for 45, I feel like i should use lhr, but i can't take the derivative of n!

Please try and explain it to me, as I already have the answers I just want to learn

 

[Oster]

25 - Have you done absolute convergence?
26 - Do you see what the graph of this function looks like?
45- Maybe you could try expanding out and then comparing it with some other sequence?

 

[Dick]

25 - Have you done absolute convergence?
26 - Do you see what the graph of this function looks like?
45- Maybe you could try expanding out and then comparing it with some other sequence?

For 25, I think the question is about convergence of the sequence, not convergence of the series. Ignore the (-1)^n for the moment and tell us whether the n/(n^2+1) part converges and to what.

 

[SammyS]

I have 4 problems left and the questions says I have to test them for converge or divergence.
Here are the problems

http://gyazo.com/f0fa5a38c5968ecb7e74103486a181bd.png

for 25, I want to take the limit as n goes to infinity however I get (-1)^\infty and stuck there

for 27, similar problem how do i take the limit of cos(\infty)

for 43 i have no idea

for 45, I feel like i should use lhr, but i can't take the derivative of n!

Please try and explain it to me, as I already have the answers I just want to learn

What methods do you know? What does it mean for a sequence to converge? ... diverge?

How did you get (-1)^∞ for 25 ? Show us your work, so we can help you.

Write out the first several terms for 27.

For 45, use logarithms.

 

[Oster]

Oops my bad. Will think before I post next time hahah. =D

 

[shemer77]

25 - Have you done absolute convergence?
26 - Do you see what the graph of this function looks like?
45- Maybe you could try expanding out and then comparing it with some other sequence?

yes, but this is a sequence I thought you can't use tests on it...?
27- i looked at the graph and it seems its going to 0 but is a graph an accurate way of knowing
45- i feel like there's a better way

For 25, I think the question is about convergence of the sequence, not convergence of the series. Ignore the (-1)^n for the moment and tell us whether the n/(n^2+1) part converges and to what.

thats going to 0

What methods do you know? What does it mean for a sequence to converge? ... diverge?

How did you get (-1)^∞ for 25 ? Show us your work, so we can help you.

Write out the first several terms for 27.

For 45, use logarithms.

I think i know almost all the methods, but these are sequences so i thought you really can't use them..
A sequence to converge means as it expands it eventually reaches a limit of a rational number, for it diverges means it goes to infinity or negative infinty

for 25 well i got the limit as n goes to infinity of that function, but if i try to take the limit it becomes
(-1)^\infty *\infty/\infty

27 - seems like its going to 0

45- I would have the ln(n!) - ln(2^n) that's infinity - infinity, how can I rewrite that?

 

[Dick]

yes, but this is a sequence I thought you can't use tests on it...?
27- i looked at the graph and it seems its going to 0 but is a graph an accurate way of knowing
45- i feel like there's a better way

thats going to 0I think i know almost all the methods, but these are sequences so i thought you really can't use them..
A sequence to converge means as it expands it eventually reaches a limit of a rational number, for it diverges means it goes to infinity or negative infinty

for 25 well i got the limit as n goes to infinity of that function, but if i try to take the limit it becomes
(-1)^\infty *\infty/\infty

27 - seems like its going to 0

45- I would have the ln(n!) - ln(2^n) that's infinity - infinity, how can I rewrite that?

This would be a lot less confusing if you work on the problems one at a time. I'd suggest you finish 25 first. -1<=(-1)^n<=1. Try using a squeeze argument. And I don't think the best answer for limit n/(n^2+1) is infinity/infinity. In fact, that's no answer at all. Didn't you also say it was 0? Why?

 

[SammyS]

... for it diverges means it goes to infinity or negative infinty
...

No, diverges simply means that it doesn't converge.

 

[SammyS]

...

45- I would have the ln(n!) - ln(2^n) that's infinity - infinity, how can I rewrite that?

Use the properties of logarithms to simplify ln(n!) - ln(2ⁿ) .

 

[shemer77]

Use the properties of logarithms to simplify ln(n!) - ln(2ⁿ) .

I don't see how that would help? can you show me the steps? wouldn't it just be ln(n!) - n*ln(2), which would still be infinity - infinity?

 

[Dick]

I don't see how that would help? can you show me the steps? wouldn't it just be ln(n!) - n*ln(2), which would still be infinity - infinity?

It's probably easier and more direct to write n!/2^n as (1/2)*(2/2)*(3/2)*(4/2)*...*(n/2). Now notice, say, that there are lots of factors that are greater than 2 if n is large. Think about a comparison test.

 

[SammyS]

Dick's idea is probably better, but here's what I had in mind.

n! = n(n-1)(n-2)...(4)(3)(2)(1).

Therefore, ln(n!) = ln(n)+ln(n-1)+ln(n-2)+...+ln(4)+ln(3)+ln(2) .

But then comparing this to (n)ln(2) can be a bit messy.

 

[Bohrok]

It's probably easier and more direct to write n!/2^n as (1/2)*(2/2)*(3/2)*(4/2)*...*(n/2). Now notice, say, that there are lots of factors that are greater than 2 if n is large. Think about a comparison test.

Wouldn't that be enough to show that the terms clearly diverge, or at least do not converge to 0?

I myself would probably go with using the ratio test.

 

[SammyS]

Wouldn't that be enough to show that the terms clearly diverge, or at least do not converge to 0?

I myself would probably go with using the ratio test.

I believe these are sequences, not series, although the only place it says that is in the title.

However, doing #45 via the logarithm does change it to being equivalent evaluating the convergence of an infinite sequence.

... which gives me an idea

Group a -ln(2) with each ln(i) to give: ln(a_n) = (ln(n)-ln(2)) + (ln(n-1)-ln(2)) + (ln(n-2)-ln(2)) +...+ ln(4)-ln(2)) + (ln(3)-ln(2)) + (ln(2)-ln(2)) + (ln(1)-ln(2)) .
​