05/29/2011, 09:02 AM

(05/29/2011, 02:25 AM)JmsNxn Wrote: I see where you're coming from, but then, why is it converging? Can't we just say:

I'm betting a is somewhere in [e, 6] range.

No, we cant say that. See, if we multiply by e^x on both sides, then the right side must be a powerseries development of ln(x)e^x at 0. This means that at least the coefficients (that are divided by n!) on the right side must be the derivatives of the function ln(x)e^x at 0. But the derivatives are all infinite while the coefficients on the right side are finite. So there will be in no way equality.

I dont know whether the right side converges or not. But even if it converges it is not equal to the left side. And if it converges for some then it is a powerseries development of some function at 0. This implies (by standard complex analysis) that it converges in an open disk around 0 of a certain radius (in the complex plane; on the real axis just in an interval (-r,r)) and outside the closed disk it can not converge. That means if it converges in [e,6] then it must converge in (-6,6).

Quote:I was a little perplexed myself when it converged, because I know that

this converges only for integer values of t. This is why I made sure to make t = 0, and not a real value, but still it makes you wonder if a derivative of the growth of something that doesn't converge will converge... but then it does, at least for x>a.

Are you sure that it doesnt converge? proof?

But if it doesnt converge then that is another problem. You take the derivative to t from a series that does not converge in a neighborhood of 0. Thats not safe.

But even then its strange that you arrive at an equation that is infinite at the left side and finite at the right side.