Prove the convergence/divergence of a given series

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Homework Statement .

Show that the series with general term ##a_n=\dfrac{1}{n^plog(n)^q}, n\geq 2##

converges if: ##q>0## and ##p>1##; ##q>1## and ##p=1##
diverges if: ##q>0## and ##p<1##; ##0<q\leq 1## and ##p=1##

The attempt at a solution.

I could solve the problem for the cases ##q>1## and ##p=1##, and ##0<q\leq 1## and ##p=1##:

For these two cases I can use the integral test where ##f(x)=\dfrac{1}{xlog(x)^q}##.

By making the substitution ##u=log(x)##, I got that ##\int_2^{+\infty} f(x)dx## converges if and only if q>1.

Now, I don't now how to show the convergence and divergence respectively for the other two remaining cases (q>0). I would appreciate any suggestions.
 

Answers and Replies

  • #2
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Do you know the Cauchy condensation test?
 
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Do you know the Cauchy condensation test?
Sure, so now I have to analyze the convergence of the series of term ##b_n=\dfrac{2^n}{(2^n)^plog(2^n)^q}##. So ##\dfrac{2^n}{(2^n)^plog(2^n)^q}=\dfrac{2^{n(1-p)}}{n^qlog(2)^q}=\dfrac{(2^{1-p})^n}{n^qlog(2)^q}##.

I know that the series ##\sum \dfrac{1}{n^q}## converges if ##q>1## and that the geometric series ##\sum (\dfrac{1}{2^{p-1}})^n## converges for ##p-1\geq 1##, which means, for ##p>1##.

Now, I have some doubts on the following:


Is it true that if ##\sum a_n## and ##\sum b_n## converge ##\implies## ##\sum a_nb_n## converges? If this is the case, by what I've previously said, the original series would converge for ##p>1## and ##q>1##. For the same reason, if what I've asked is true, then for ##q=1## and ##p>1## the original series converges as well.

I still don't know what to do if ##q>0## and ##p<1##
 
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  • #4
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Sure, so now I have to analyze the convergence of the series of term ##b_n=\dfrac{2^n}{(2^n)^plog(2^n)^q}##. So ##\dfrac{2^n}{(2^n)^plog(2^n)^q}=\dfrac{2^{n(1-p)}}{n^qlog(2)^q}=\dfrac{(2^{1-p})^n}{n^qlog(2)^q}##.

I know that the series ##\sum \dfrac{1}{n^q}## converges if ##q>1## and that the geometric series ##\sum (\dfrac{1}{2^{p-1}})^n## converges for ##p-1\geq 1##, which means, for ##p>1##.

Now, I have some doubts on the following:


Is it true that if ##\sum a_n## and ##\sum b_n## converge ##\implies## ##\sum a_nb_n## converges?
In this case (series of positive numbers) it is true. But you could just apply the comparison criterion instead of this theorem, which I guess you will need to prove in order to use it (you can prove it by the comparison criterion!)

If this is the case, by what I've previously said, the original series would converge for ##p>1## and ##q>1##.
Try the comparison test.

I still don't know what to do for the cases ##q=1## and ##p>1##
Comparison test will work here too. The idea is that the exponential will always dominate the polynomial factor. So if the exponential goes to ##0##, then it will do so way faster than any polynomial factor, making the series diverge.

##q>0## and ##p<1##
The exponential factor will diverge here and this will again dominate the entire series. So try to show divergence. Think about the limit of the sequence of terms.
 
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In this case (series of positive numbers) it is true. But you could just apply the comparison criterion instead of this theorem, which I guess you will need to prove in order to use it (you can prove it by the comparison criterion!)



Try the comparison test.



Comparison test will work here too. The idea is that the exponential will always dominate the polynomial factor. So if the exponential goes to ##0##, then it will do so way faster than any polynomial factor, making the series diverge.



The exponential factor will diverge here and this will again dominate the entire series. So try to show divergence. Think about the limit of the sequence of terms.

For convergence I could prove with the comparison test, as you've suggested. For q>0 and p<1, I could show divergence by the ratio test, thanks for the help!
 

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