For what values of p and q does a series converge?

  • #1

Homework Statement


For what values of p and q does [tex]\sum\limits_{n=2}^\infty \frac{1}{n^q\ln(n)^q}[/tex] converge?



The Attempt at a Solution


I've tried a couple of tests, but given that there are two variables (p and q), I'm not really sure how to proceed. My hunch is that I have to use the integral test, which says that the series will converge if [tex]\int^{\infty}_2 \frac{1}{n^q\ln(n)^q}\,dn[/tex] converges, but this isn't something I can integrate nicely, so I'm not quite sure.

Could someone point me in the right direction?
 

Answers and Replies

  • #2
88
1
I think you've miss typed somewhere. Your sum only has a q, not a p.
 
  • #3
STEMucator
Homework Helper
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Homework Statement


For what values of p and q does [tex]\sum\limits_{n=2}^\infty \frac{1}{n^q\ln(n)^q}[/tex] converge?



The Attempt at a Solution


I've tried a couple of tests, but given that there are two variables (p and q), I'm not really sure how to proceed. My hunch is that I have to use the integral test, which says that the series will converge if [tex]\int^{\infty}_2 \frac{1}{n^q\ln(n)^q}\,dn[/tex] converges, but this isn't something I can integrate nicely, so I'm not quite sure.

Could someone point me in the right direction?

Hint : Try considering what's going on around ##q<0##. Try ##q=-1## for example.

How about when ##0 \le q \le 1##?

What can you conclude?
 
  • #4
Oh, Sorry! It's actually
For what values of p and q does [tex]\sum\limits_{n=2}^\infty \frac{1}{n^p\ln(n)^q}[/tex] converge?
 
  • #5
STEMucator
Homework Helper
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Oh, Sorry! It's actually
For what values of p and q does [tex]\sum\limits_{n=2}^\infty \frac{1}{n^p\ln(n)^q}[/tex] converge?

It's all good. I wish it was just ##q##, would've made it a bit easier.

As for ##p## and ##q## though. Do you know anything about the comparison test?
 
  • #6
Yeah! I couldn't think of anything useful to compare it to, though.
 
  • #7
pasmith
Homework Helper
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Hint: if [itex]q > 0[/itex] and [itex]\ln n > 1[/itex], then [itex]\displaystyle\frac{1}{n^p (\ln n)^q} < \frac{1}{n^p}[/itex].

Similarly, if [itex]p > 1[/itex] and [itex]n > 1[/itex] then [itex]\displaystyle\frac{1}{n^p (\ln n)^q} <
\frac{1}{n (\ln n)^q}[/itex].

Can you see how to integrate [itex]\displaystyle\int_2^\infty \frac{1}{x (\ln x)^q}\,\mathrm{d}x[/itex] by substitution?
 
  • #8
STEMucator
Homework Helper
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Yeah! I couldn't think of anything useful to compare it to, though.

Alright, I'll help you run through one case since there's a couple cases to consider.

The series at hand : ##\sum\limits_{n=2}^\infty \frac{1}{n^p\ln(n)^q}##

Ask yourself, which term in the denominator of your series is your series more dependent on? That is, which function grows faster? ##n## or ##ln(n)##? Forget about ##p## and ##q## for a moment.

Without too much thought it's easy to see ##n>ln(n)## for sufficiently large ##n## independent of ##p## and ##q## and you can use that to your advantage when comparing.

##\sum\limits_{n=2}^\infty \frac{1}{n^p\ln(n)^q} \le \sum\limits_{n=2}^\infty \frac{1}{n^p}##

So for all values of ##q##, the convergence of the series on the left depends on the convergence of the series on the right. What do you know about the series on the right? What can you conclude from all this? This will give you your first case.
 

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