05/29/2011, 02:25 AM
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.
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.
Perhaps I'm making an error somewhere and the convergence is right for the wrong reasons.
\( ln(x) = e^{-x}(\sum_{n=0}^{\infty} \frac{x^n}{n!}\psi_0(n+1))\,\,\,\{x| x > a > 0, x,a \in \R\} \) I'm betting a is somewhere in [e, 6] range.
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.
Perhaps I'm making an error somewhere and the convergence is right for the wrong reasons.
