1. ## Convergent/Divergent Sequences...

Which of the following sequences are convergent and which are divergent?
a) $a_{n} = (1+2n)^{\frac{1}{n}}$
b) $a_{n} = cos(n\pi)$
c) $a_{n} = \frac{cos(3n)}{(1+n^{1/2})}$

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,

2. (a) Diverge. The numerator is always larger than the denominator.

3. 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.
So is it correct to say that the sequence diverges?
The correct conclusion is that the sequence converges .

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.
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?

4. If the OP meant $(1+2n)^{(1/n)}$ then yes the sequence converges.

5. ## more confused...

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

6. Please write as (1+2n)^(1/2) if you meant it to be a square root. This might also be your source of error.

7. ## yes...

yes that is correct...
i did mean (1+2n)^(1/2) ....
though workin it by hand.. there were no errors...
any suggestionss??
zee

8. 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.
$
\left|\frac{\cos(3n)}{1+\sqrt{n}}\right| \le \frac{1}{1+\sqrt{n}} \le \frac{1}{\sqrt{n}}
$

Therefore
$
\lim_{n \to \infty} \frac{\cos(3n)}{1+\sqrt{n}} = 0
$

by the squeeze theorem. This is a limit of a sequence.

However
$\sum_{n=1}^\infty \frac{1}{\sqrt{n}}
$

does not converge (p-series). This is an infinite series .

Check what your original question was. I suspect it was about limits of a sequence. So you shouldn't be thinking in terms of p-series in this context.
I guess that is why you are confused.

This does not, however, settle whether $\sum_{n=1}^\infty \frac{\cos(3n)}{1+\sqrt{n}}$ converges or diverges. However, it cannot converge absolutely. I suspect it converges conditionally - anybody ready to prove this?