Question

I need help finding the P values for which this series converges.

Answer 1

\[\sum_{2}^{infinity}1/n(lnn)^P\]

Answer 2

I thought it was just for any P>1 but assuming thats too obvious

Answer 3

let's try taking ratio test.
http://en.wikipedia.org/wiki/Ratio_test

\[ \lim_{n->\infty} \frac{\frac{1}{(n+1)(\ln(1+n))^p}}{\frac{1}{n (\ln(n))^p}} < 1 \]

Answer 4

\[ \lim_{n->\infty} \frac{n (\ln(n))^p}{(n+1) (\ln(n+1))^p } < 1\]
\[ \lim_{n->\infty} \frac{(\ln(n))^p}{(\ln(n+1))^p } < \frac{n+1}{n}\]
\[ \lim_{n->\infty} (\frac{\ln(n)}{\ln(n+1) })^p < \frac{n+1}{n}\]
\[ \lim_{n->\infty} (\frac{\ln(n)}{\ln(n+1) })^p < 1 +\frac{1}{n}\]
\[ \lim_{n->\infty} \frac{\ln(n)}{\ln(n+1) } < (1 +\frac{1}{n})^{1/p}\]

Answer 5

http://www.wolframalpha.com/input/?i=lim+n-%3Einf+ln%28n%29%2Fln%281%2Bn%29

the rhs is (1+1/inf)^(1/p) = (1)^(1/p)

or 1 < 1 for p>1 ... doesn't make any sense.

Answer 6

*p > 0

Answer 7

seems to diverge
http://www.wolframalpha.com/input/?i=lim+n-%3Einf+sum%5B1%2F%28n+%28ln+n%29%5E0.003%29%2C+2%2C+n%5D

Answer 8

seems to diverge to this value
http://www.wolframalpha.com/input/?i=lim+n-%3Einf+sum%5B1%2F%28n+%28ln+n%29%5En%29%2C+2%2C+n%5D

Answer 9

it converges for p>1

Answer 10

use the integral test

Answer 11

here I'm assuming your sum is

\[\sum_{n=2}^{\infty}\frac{1}{n[\ln(n)^p]}\]

Answer 12

Let
\[ f(x) =\frac 1 {(x \ln x)^p}
\]
\[
\int_2^{c} f(x) dx= \frac{\ln ^{1-p}(2)-\ln
^{1-p}(c)}{p-1}
\]
The last expression has a finite limit when c goes to Infinity if p>1
Use the intgral test and conclude that p>1

Answer 13

seems so ... but why did my ratio test failed?? How do I know when to use integral test??

Answer 14

Sorry f is

\[f(x) =\frac 1 {x( \ln x)^p}
\]
The rest is fine above.

Answer 15

You keep tryimg until one works.

Answer 16

the ratio and root tests don't always work...they can be inconclusive.

Answer 17

Hahah ... if it fails (must diverge) ... or I get stupid logic like i got above (1 < 1^(1/p))??

Answer 18

if the test is inconclusive then the sum could be finite or it could diverge

for example

\[\sum_{n=1}^{\infty}\frac{1}{n^2+1}\] and
\[\sum_{n=1}^{\infty}\frac{1}{n+1}\]


the ratio test is inconclusive in both cases but the first one converges and the second diverges.

Answer 19

Thanks to you both ... it improved my understanding.