Convergent/Divergent Sequences...

• May 20th 2005, 11:19 PM
zee
Convergent/Divergent Sequences...
Which of the following sequences are convergent and which are divergent?
a) $\displaystyle a_{n} = (1+2n)^{\frac{1}{n}}$
b) $\displaystyle a_{n} = cos(n\pi)$
c) $\displaystyle a_{n} = \frac{cos(3n)}{(1+n^{1/2})}$

This is what I got:
a) By the squeeze theorem, 1 < (1+2n)^1/2 < 3
and the sequence decreases as a(subscript n) > a(subscript n+1)
So is it correct to say that the sequence diverges?

b) {a(subscript n)} = {cosPI, cos2PI, cos3PI, cos4PI...}
= {-1,1,-1,1,...}
Since this sequence oscillates between -1 and 1, so it diverges.
Is this correct?

c) since cos(3n) <= 1 for all n, we have
cos(3n)/(1+n^1/2) <= 1/(1+n^1/2) <= 1/(n^1/2)
and we know that 1/n^1/2 is divergent (since p=1/2)
Is this correct?

I am almost sure that b and c are correct, but as for question a,
• May 21st 2005, 06:54 AM
paultwang
(a) Diverge. The numerator is always larger than the denominator.
• May 21st 2005, 12:10 PM
hpe
Quote:

Originally Posted by zee
Which of the following sequences are convergent and which are divergent?
a) a(subscript n) = (1+2n)^1/n
b) a(subscript n) = cos(nPI)
c) a(subscript n) = cos(3n)/(1+n^1/2)

This is what I got:
a) By the squeeze theorem, 1 < (1+2n)^1/2 < 3
and the sequence decreases as a(subscript n) > a(subscript n+1)

You need to prove that.
Quote:

So is it correct to say that the sequence diverges?
The correct conclusion is that the sequence converges .

Quote:

b) {a(subscript n)} = {cosPI, cos2PI, cos3PI, cos4PI...}
= {-1,1,-1,1,...}
Since this sequence oscillates between -1 and 1, so it diverges.
Is this correct?
Yes. But check if you need a more formal argument.
Quote:

c) since cos(3n) <= 1 for all n, we have
cos(3n)/(1+n^1/2) <= 1/(1+n^1/2) <= 1/(n^1/2)
and we know that 1/n^1/2 is divergent (since p=1/2)
Is this correct?
Incorrect. This is a sequence, not a series. Use the squeeze theorem.

Btw, would you please confirm here in writing that these are not questions from a take-home final?
• May 21st 2005, 02:19 PM
paultwang
If the OP meant $\displaystyle (1+2n)^{(1/n)}$ then yes the sequence converges.
• May 26th 2005, 03:19 AM
zee
more confused...
:eek:
I dont understand why (1+2n)^1/n is convergent.
I did apply the squeeze theorem as above
1 < (1+2n)^1/2 < 3
but still it turns out to be divergent because the
series is decreasing. I even plotted a table to check
if it was decreasing. What am I doing wrong?

Also for the other question, cos(3n)/(1+n^1/2):
I applied the squeeze theorem here too.
cos(3n)/(1+n^1/2) <= 1/(1+n^1/2) <= 1/(n^1/2)
as above. And the rule is:
For: 1/n^p, if p>1, it is convergent, if p<=1, it's divergent.
In this question it is clearly divergent as p = 1/2.

I know i must be dong something wrong as I'm getting
the answers for all the practice questions to be divergent.
I do have all the criterias for when they are div and conv,
but somehow the end answers always satisfy the div property.
zee
• May 26th 2005, 06:22 AM
paultwang
Please write as (1+2n)^(1/2) if you meant it to be a square root. This might also be your source of error.
• May 27th 2005, 11:07 AM
zee
yes...
yes that is correct...
i did mean (1+2n)^(1/2) ....
though workin it by hand.. there were no errors...
any suggestionss??
zee
• Jun 1st 2005, 07:00 PM
hpe
Quote:

Also for the other question, cos(3n)/(1+n^1/2):
I applied the squeeze theorem here too.
cos(3n)/(1+n^1/2) <= 1/(1+n^1/2) <= 1/(n^1/2)
as above. And the rule is:
For: 1/n^p, if p>1, it is convergent, if p<=1, it's divergent.
In this question it is clearly divergent as p = 1/2.
$\displaystyle \left|\frac{\cos(3n)}{1+\sqrt{n}}\right| \le \frac{1}{1+\sqrt{n}} \le \frac{1}{\sqrt{n}}$
$\displaystyle \lim_{n \to \infty} \frac{\cos(3n)}{1+\sqrt{n}} = 0$
$\displaystyle \sum_{n=1}^\infty \frac{1}{\sqrt{n}}$
This does not, however, settle whether $\displaystyle \sum_{n=1}^\infty \frac{\cos(3n)}{1+\sqrt{n}}$ converges or diverges. However, it cannot converge absolutely. I suspect it converges conditionally - anybody ready to prove this?