Another Question for a P-Series Problem

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Edder

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Determine all values of p for which the series is convergent, and express answer in interval notation.

\(\displaystyle \sum_{n=2}^{\infty} (-1)^{n-1} \frac{ln(n)}{2n}^{p}\)

How would I find the values of p that makes this series a convergent one. I know that its an alternating series, so would I just focus on \(\displaystyle \frac{ln(n)}{2n}^{p}\) ?
I am just not quite sure how to go about this problem.

Any feedback and help appreciated, thanks.
 
Alternating...How can you ensure that the terms are decreasing toward zero? What will a ratio test tell you?
 
Edder said:
Determine all values of p for which the series is convergent, and express answer in interval notation.
\(\displaystyle \sum_{n=2}^{\infty} (-1)^{n-1} \frac{ln(n)}{2n}^{p}\)
I just focus on \(\displaystyle \frac{ln(n)}{2n}^{p}\)
Click to expand...
It appears from looking at the LaTeX that you meant \(\displaystyle \left(\frac{ln(n)}{2n}\right)^{p}\).
Is that the case? If so it is not what appears.
If not, what do you mean?
 
pka said:
It appears from looking at the LaTeX that you meant \(\displaystyle \left(\frac{ln(n)}{2n}\right)^{p}\).
Is that the case? If so it is not what appears.
If not, what do you mean?
Click to expand...

No, only the ln(n) is raised to the p power. I still can't quite figure this problem out. I don't think that I would the ratio test, as my professor hasn't taught us this method yet. Is there a different method to solve this problem? So far we have covered the integral, comparison, and absolute divergent tests.
 
At last - information! This is good.

Ignoring the 'p', can you determine if the successive term are decreasing?

Ignoring the 'p', can you determine the maximum term.
 
tkhunny said:
At last - information! This is good.

Ignoring the 'p', can you determine if the successive term are decreasing?

Ignoring the 'p', can you determine the maximum term.
Click to expand...

If I ignore the p, then yes the \(\displaystyle \frac{ln(n)}{2n}\) is a decreasing series. Then the limit as n goes to infinity is equal to zero. Thus, the series passes the test of an alternating series. But the p has to make some kind of difference, right? As the p gets bigger, the numerator will outgrow the denominator.

So, would the p have to equal zero to to make this series convergent?
 
One step at a time. NOT ignoring the 'p', are the succesive terms decreasing with p = 0? Absolutely not. Don't get all jumpy.

Why do you believe it is a decreasing sequence? What proof have you?

Next, answer the other question. If the terms are ALWAYS less than 1, we just agreed to all p >= 1. Why?
 
tkhunny said:
One step at a time. NOT ignoring the 'p', are the succesive terms decreasing with p = 0? Absolutely not. Don't get all jumpy.

Why do you believe it is a decreasing sequence? What proof have you?

Next, answer the other question. If the terms are ALWAYS less than 1, we just agreed to all p >= 1. Why?
Click to expand...


I took the first derivative test and found \(\displaystyle \frac{ln(n)}{n}\) to be decreasing from 2 to infinity. So therefor, the bn by the alternating series test should be convergent. But if p=0, then no it is not decreasing as it is then a harmonic series of \(\displaystyle \frac{1}{n}\)

Then for this series to be convergent, p would have to be from (0, infinity), right?
 
Let me describe how you are thinking.

"I have one additional piece of information. I should now be able to see the answer."

This is not sufficient thinking. Sometimes things come in little chunks. You must process them all until you understand it all.

Let's just start on the left. p < 0

\(\displaystyle \frac{\left[ln(n)\right]^{p}}{2n}\)

This can be rewritten:

\(\displaystyle \frac{1}{2n\cdot \left[ln(n)\right]^{somethingpositive}}\)

I'm feeling pretty good about p < 0. Can you PROVE it?

Let's try p = 0.

You wisely observed that this produces 1/(2n) alternating. Does that converge?

Now for p > 0...

Isn't that where we started?

Sit down and think on the problem. Ponder it's entire Domain, in pieces if necessary. Keep working until you are sure you see what is going on. Don't just jump at it every time you learn something.
 
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