For which positive real numbers a does the series converge


by .d9n.
Tags: converge, numbers, positive, real, series
.d9n.
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#19
Feb1-12, 09:24 AM
P: 61
if r>1 which would make -p=r, what values for a would make log(a) negative?
Dick
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#20
Feb1-12, 09:30 AM
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Quote Quote by .d9n. View Post
does it converge if p<-1
That sounds good. So you want log(a)<(-1), right? Solve that for a. What's the inverse function to a log? And I don't think they mean log to the base 10.
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#21
Feb1-12, 09:49 AM
P: 61
so -1<a<1 a=/0,
say y=log(a) then the inverse is a=e^y ?
Dick
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#22
Feb1-12, 10:00 AM
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Quote Quote by .d9n. View Post
so -1<a<1 a=/0,
say y=log(a) then the inverse is a=e^y ?
No again to the first line. But yes to the second. e^(log(a))=a. Take log(a)<(-1) and exponentiate both sides of the inequality.
.d9n.
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#23
Feb1-12, 10:05 AM
P: 61
so a<e^-1= a<0.367879441?
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#24
Feb1-12, 10:08 AM
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Quote Quote by .d9n. View Post
so a<e^-1= a<0.367879441?
Now you've got it. And you also need 0<a. Otherwise the log (or your original series) isn't even defined.
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#25
Feb1-12, 10:08 AM
P: 61
so sum n=1 to infinity of a^log(n) converges when 0<a<0.3678...
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#26
Feb1-12, 10:18 AM
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Quote Quote by .d9n. View Post
so sum n=1 to infinity of n^log(a) converges when 0<a<0.3678...
Yes, but I'd write e^(-1) or 1/e instead of 0.3678... There's no need to replace an exact number with an approximation.
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#27
Feb1-12, 10:20 AM
P: 61
ok, brilliant. thanks for all your help, much appreciated


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