Andrey Kulsha wrote:

> The series

Sorry, there was a typo: k goes from 2 to infinity.

>

> sum(1/(klogk),k,1,+infinity)

>

> diverges.

Phil Carmody wrote:

> OK, that does seem to make sense. Thanks.

It converges for any n>1.

> (Am I right in thinking that the sum of 1/(k.log(k)^n)

> would diverge for any constant n?)

You may calculate an integral(f(k)*dk,C,+infinity) to check

the series

sum(f(k),k,C,+infinity),

if f(k) is smooth for C<=k<+infinity.

Best wishes,

Andrey