How to tell which values of $p$ let this sum converge
This is the sum:
$$sumlimits_{n=3}^inftyfrac{1}{ncdotln(n)cdotln(ln(n))^p}$$
How do I tell which values of $p$ allow this to converge? The ratio test isn’t working out for me at all.
sequences-and-series convergence
edited Nov 20 at 3:00
T. Bongers
22.8k54661
asked Nov 20 at 2:47
Maddie
162
add a comment |
This is the sum:
$$sumlimits_{n=3}^inftyfrac{1}{ncdotln(n)cdotln(ln(n))^p}$$
How do I tell which values of $p$ allow this to converge? The ratio test isn’t working out for me at all.
sequences-and-series convergence
edited Nov 20 at 3:00
T. Bongers
22.8k54661
asked Nov 20 at 2:47
Maddie
162
This looks like a case for the en.wikipedia.org/wiki/Cauchy_condensation_test . But are you sure that it is $ln(ln(n))^p$ instead of $(lnln(n))^p$?
– trancelocation
Nov 20 at 3:08
add a comment |
This is the sum:
$$sumlimits_{n=3}^inftyfrac{1}{ncdotln(n)cdotln(ln(n))^p}$$
How do I tell which values of $p$ allow this to converge? The ratio test isn’t working out for me at all.
sequences-and-series convergence
edited Nov 20 at 3:00
T. Bongers
22.8k54661
asked Nov 20 at 2:47
Maddie
162
This is the sum:
$$sumlimits_{n=3}^inftyfrac{1}{ncdotln(n)cdotln(ln(n))^p}$$
How do I tell which values of $p$ allow this to converge? The ratio test isn’t working out for me at all.
sequences-and-series convergence
sequences-and-series convergence
edited Nov 20 at 3:00
T. Bongers
22.8k54661
asked Nov 20 at 2:47
Maddie
162
edited Nov 20 at 3:00
T. Bongers
22.8k54661
asked Nov 20 at 2:47
Maddie
162
edited Nov 20 at 3:00
T. Bongers
22.8k54661
edited Nov 20 at 3:00
T. Bongers
22.8k54661
edited Nov 20 at 3:00
T. Bongers
22.8k54661
22.8k54661
asked Nov 20 at 2:47
Maddie
162
asked Nov 20 at 2:47
Maddie
162
asked Nov 20 at 2:47
Maddie
162
162
This looks like a case for the en.wikipedia.org/wiki/Cauchy_condensation_test . But are you sure that it is $ln(ln(n))^p$ instead of $(lnln(n))^p$?
– trancelocation
Nov 20 at 3:08
add a comment |
This looks like a case for the en.wikipedia.org/wiki/Cauchy_condensation_test . But are you sure that it is $ln(ln(n))^p$ instead of $(lnln(n))^p$?
– trancelocation
Nov 20 at 3:08
This looks like a case for the en.wikipedia.org/wiki/Cauchy_condensation_test . But are you sure that it is $ln(ln(n))^p$ instead of $(lnln(n))^p$?
– trancelocation
Nov 20 at 3:08
This looks like a case for the en.wikipedia.org/wiki/Cauchy_condensation_test . But are you sure that it is $ln(ln(n))^p$ instead of $(lnln(n))^p$?
– trancelocation
Nov 20 at 3:08
add a comment |
1 Answer
1
active
oldest
votes
The ratio test will not work because this series converges/diverges far too slowly for the test to give any information. Rather, the integral test is most useful when studying things related to $p$-series. In particular, if we set $u = ln ln x$ then we can find that
begin{align*}
int_3^{infty} frac{1}{x ln x (ln ln x)^p} , dx &= int_{ln ln 3}^{infty} frac{1}{u^p} , du
end{align*}
which is now a vastly easier integral to study.
answered Nov 20 at 3:00
T. Bongers
22.8k54661
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The ratio test will not work because this series converges/diverges far too slowly for the test to give any information. Rather, the integral test is most useful when studying things related to $p$-series. In particular, if we set $u = ln ln x$ then we can find that
begin{align*}
int_3^{infty} frac{1}{x ln x (ln ln x)^p} , dx &= int_{ln ln 3}^{infty} frac{1}{u^p} , du
end{align*}
which is now a vastly easier integral to study.
answered Nov 20 at 3:00
T. Bongers
22.8k54661
add a comment |
The ratio test will not work because this series converges/diverges far too slowly for the test to give any information. Rather, the integral test is most useful when studying things related to $p$-series. In particular, if we set $u = ln ln x$ then we can find that
begin{align*}
int_3^{infty} frac{1}{x ln x (ln ln x)^p} , dx &= int_{ln ln 3}^{infty} frac{1}{u^p} , du
end{align*}
which is now a vastly easier integral to study.
answered Nov 20 at 3:00
T. Bongers
22.8k54661
add a comment |
The ratio test will not work because this series converges/diverges far too slowly for the test to give any information. Rather, the integral test is most useful when studying things related to $p$-series. In particular, if we set $u = ln ln x$ then we can find that
begin{align*}
int_3^{infty} frac{1}{x ln x (ln ln x)^p} , dx &= int_{ln ln 3}^{infty} frac{1}{u^p} , du
end{align*}
which is now a vastly easier integral to study.
answered Nov 20 at 3:00
T. Bongers
22.8k54661
The ratio test will not work because this series converges/diverges far too slowly for the test to give any information. Rather, the integral test is most useful when studying things related to $p$-series. In particular, if we set $u = ln ln x$ then we can find that
begin{align*}
int_3^{infty} frac{1}{x ln x (ln ln x)^p} , dx &= int_{ln ln 3}^{infty} frac{1}{u^p} , du
end{align*}
which is now a vastly easier integral to study.
answered Nov 20 at 3:00
T. Bongers
22.8k54661
answered Nov 20 at 3:00
T. Bongers
22.8k54661
answered Nov 20 at 3:00
T. Bongers
22.8k54661
answered Nov 20 at 3:00
T. Bongers
22.8k54661
22.8k54661
add a comment |
add a comment |
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This looks like a case for the en.wikipedia.org/wiki/Cauchy_condensation_test . But are you sure that it is $ln(ln(n))^p$ instead of $(lnln(n))^p$?
– trancelocation
Nov 20 at 3:08