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 What is the convergence radius of this power series? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 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: $ln(x) = e^{-x}(\sum_{n=0}^{\infty} \frac{x^n}{n!}\psi_0(n+1))\,\,\,\{x| x > a > 0, x,a \in \R\}$ 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 $x\neq 0$ 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 $\frac{d^t}{dx^t}e^x = e^x = \sum_{n=0}^{\infty} \frac{x^{n-t}}{\Gamma(n+1-t)}$ 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. « Next Oldest | Next Newest »