For which positive real numbers a does the series converge

.d9n.

so it converges when r>1?

.d9n.

if r>1 which would make -p=r, what values for a would make log(a) negative?

Dick

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.

.d9n.

so -1<a<1 a=/0,
say y=log(a) then the inverse is a=e^y ?

Dick

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.

so a<e^-1= a<0.367879441?

Dick

Now you've got it. And you also need 0<a. Otherwise the log (or your original series) isn't even defined.

.d9n.

so sum n=1 to infinity of a^log(n) converges when 0<a<0.3678...

Dick

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.

.d9n.

ok, brilliant. thanks for all your help, much appreciated