# Math Help - Find all positive values of b for which the following series is convergent.

1. ## Find all positive values of b for which the following series is convergent.

hi

sigma from to infinity for b^(ln n)

hmmm i tried to write :
b^(ln n) = [ ( b^(ln n) )^1/n ] ^ n
and deal with it as an goemetric series .. but it failed ;p

help :/

2. Originally Posted by TWiX
hi

sigma from to infinity for b^(ln n)

hmmm i tried to write :
b^(ln n) = [ ( b^(ln n) )^1/n ] ^ n
and deal with it as an goemetric series .. but it failed ;p

help :/
$0 \leq b < e^{-1}$

3. Can you give me its idea ?

4. Originally Posted by TWiX
Can you give me its idea ?
Apply the ratio test and force the limit to be less than 1 to get the radius of convergence. Then test the endpoints to get the interval of convergence.

5. Is this solution work here ?

b^(ln n) = [ e^(ln b) ]^(ln n) = [ e^(ln n) ]^(ln b) = n^(ln b) = 1/[ n^(ln(1/b)) ]
which is a p-series.

ln(1/b) > 1 ----> 1/b > e ----> 0 < b < 1/e

6. Originally Posted by mr fantastic
Apply the ratio test and force the limit to be less than 1 to get the radius of convergence. Then test the endpoints to get the interval of convergence.
Edit:
your solution is wrong
the limit is 1
test failed
i noticed this now.

7. Originally Posted by TWiX
Is this solution work here ?

b^(ln n) = [ e^(ln b) ]^(ln n) = [ e^(ln n) ]^(ln b) = n^(ln b) = 1/[ n^(ln(1/b)) ]
which is a p-series.

ln(1/b) > 1 ----> 1/b > e ----> 0 < b < 1/e
Yes, that's a very good solution