Question

Help Please!

Answer 1

Best results are obtained by displaying the complete problem and showing your best work.

Answer 2

Determine whether the sequence diverges or converges.

\[a_n = \frac{ \cos^2n }{ 2^n }\]

Answer 3

@tkhunny Woooooooow. Thanks. xD

Answer 4

First, give an impression.

Answer 5

I am thinking it Converges, since the numerator Cos is limited within [-1,1]. And the denominator grows rapidly to infinity.

Answer 6

"grows rapidly" is too vague. Also, it's cos^2(n), so it's really [0,1].

2nd impression after counsel?

Answer 7

Oh I see. Well since the numerator is bounded, and the denominator grows infinitely. This sequence approaches 0.

Answer 8

Correct. Written more formally, \(0 \le \dfrac{\cos^{2}(n)}{2^n} \le \dfrac{1}{2^n}\). The last expression's sequence is geometric with ration 1/2 and always converges.

Answer 9

Oh okay I see. :)

Answer 10

Good work. Only two impressions to a complete solution. VERY good work.

Answer 11

ration lolol
I make that typo constantly :)

Answer 12

Thanks!
How about for...

\[a_n = \left( 1+\frac{ 1 }{ n } \right)^n \]
\[a_n = \left( 1+\frac{ 1 }{ n } \right)^n \neq 1^n+\frac{ 1 }{ n }^n\]

Answer 13

I think just about everyone does. I almost feel like putting it in my filters, but sometimes I want to say "ration".

You should get away from phrases like "grows infinitely" or "approaches infinity". Generally, "increases without bound" is more descriptive. "infinity" isn't really a place.

Answer 14

Nope, that's a trickier one. You need to introduce a logarithm to discern it correctly.

Answer 15

Would that be that ...\[\frac{ 1 }{ n }\]
approaches 0. Thus we can look at it as \[(1+0)^n = (1)^n = 1 (Converges)\]

Answer 16

No good. Look at the sequence \(\log(a_{n})\)

Answer 17

Um. Sorry, I dont know what \[\log(a_n) \] is D:

Answer 18

\(\log(a_{n}) = \log\left(1+\dfrac{1}{n}\right)^{n} = n\cdot\log\left(1+\dfrac{1}{n}\right) = \dfrac{\log\left(1+\dfrac{1}{n}\right)}{\dfrac{1}{n}}\)

What say you of the convergence of that thing?

Answer 19

Well...\[\frac{ \log(1+\frac{ 1 }{ n } )}{ \frac{ 1 }{ n }}=\frac{ \log(1) \times \log(\frac{ 1 }{ n }) }{ \frac{ 1 }{ n } }=\frac{ 0 \times und }{ 0 } = 0\]

Answer 20

No good. Why do you think 0/0 = 0? It is called an "Indeterminate Form". Can you find that limit?

Answer 21

Ooops. Right right.

Answer 22

Really, \(\log(1 + 1/n) \ne log(1)\cdot\log(1/n)\). I should fail you for the whole semester for that. :-)

Answer 23

So we can use l'hospitals?
\[\frac{ \log(1+\frac{ 1 }{ n }) }{ \frac{ 1 }{ n } }=\frac{ \frac{ 1 }{ 1+\frac{ 1 }{ n } } }{ 1 } = 1+\frac{ 1 }{ n } = 1+0 = 1 (Converges)\]

Answer 24

@tkhunny Oh wow really?
I thought log rules states that\[\log(a+b) = \log(a) \times \log(b)\]

Answer 25

Your log rule is backwards.

EXCELLENT!!!!!! So, if \(\log(a_{n}) \rightarrow 1\), what does that say about \(a_{n}\)?

Answer 26

That it converges?

Answer 27

...to what? log(what) = 1?

Answer 28

OH.\[\log(a_n)=1 \rightarrow a_n = e^n (Converges)\]

Answer 29

Get rid of the n and you are there.

Answer 30

O right right. Haha typo.

Answer 31

I expect more than half your class will miss this one.
If you have the same variable in the exponent and out of the exponent, please think about logarithmic pursuits.

Answer 32

Okay thank you very much for the help! I really appreciate it.

Answer 33

No worries. It's what we do. Stay awesome.