For what p does this converge?

[fk378]

Homework Statement

Given the limit (B-->inf) [(lnB)^(-p+1)] / (-p+1) - [(ln1)^(-p+1)] / (-p+1)

For what p does this converge?

The Attempt at a Solution

For the left side of the minus sign,
if 1-p<0 --> 0
if 1-p>0 --> inf

but for the ln(1) on the right side of the minus sign, ln(1)=0, so it would be
if 1-p<0 then you would get a zero in the denominator (since ln(1) would be raised to a negative power) and therefore ---> inf
if 1-p>0 then you would have zero as the numerator, but be left with infinity as the other value and hence ---> inf

The book says that for p>1 then it will converge. But how can that be, if you'll end up with a negative exponent for the 0??

 

[HallsofIvy]

Is that really "ln(1)" in the last term? ln(1)= 0 of course. Also parentheses would help.
Finally, you should not use "B" and "b" to mean the same thing but since there is no "b" in function I am going to assume you are. I am also going to assume that is [ln(b^(-p+1))]/[-p+1] - [(ln1)^(-p+1)] / [-p+1] which, since ln(1)= 0 is the same as ln(B^-p+1/(-p+1)= (-p+1)ln(B)/(-p+1)= ln(b). So you have [itex]\lim{b\rightarrow \infty} ln(b) which does not converge for any p.

If you meant something else please clarify.

That converges to ln(B

 

[fk378]

Is that really "ln(1)" in the last term? ln(1)= 0 of course. Also parentheses would help.
Finally, you should not use "B" and "b" to mean the same thing but since there is no "b" in function I am going to assume you are. I am also going to assume that is [ln(b^(-p+1))]/[-p+1] - [(ln1)^(-p+1)] / [-p+1] which, since ln(1)= 0 is the same as ln(B^-p+1/(-p+1)= (-p+1)ln(B)/(-p+1)= ln(b). So you have [itex]\lim{b\rightarrow \infty} ln(b) which does not converge for any p.

If you meant something else please clarify.

That converges to ln(B

Please do not assume that when I stated (lnB)^(-p+1) I meant ln(B^-p+1) because I did not. The exponent is for the entire natural log expression. Therefore your statement regarding the answer to be just lnB is invalid. Thank you though.