(05/27/2011, 09:00 AM)bo198214 Wrote: I guess the problem in your derivations occurs after this line:
(05/26/2011, 02:50 AM)JmsNxn Wrote: \( 0 = \sum_{n=0}^{\infty} x^{n-t}\frac{\Gamma(n+1)}{n!\Gamma(n+1-t)}(\psi_0(n+1-t) - \ln(x)) \)
I assume this line is still convergent, however if you separate the difference into two sides you work with two divergent series.
Well the next line must be true, because, using pari gp
\( \ln(x) = e^{-x} (\sum_{n=0}^{\infty} \frac{x^n}{n!} \psi_0(n+1)) \)
this series converges, at least for values like ln(20), ln(100), and numbers just greater than e they have very close convergence, maybe 4 to 6 decimal places.
It was this series \( ln(x) = \sum_{n=0}^{\infty} x^n (\sum_{k=0}^{n} (-1)^k \frac{\sum_{c=1}^{n-k}\frac{1}{c} - \gamma}{k!(n-k)!}) \) that doesn't converge at all.
(05/27/2011, 09:00 AM)bo198214 Wrote: Remember \( \lim_{n\to\infty} a_n - b_n = \lim_{n\to\infty} a_n - \lim_{n\to\infty} b_n \) *only* if all (or at least two of the three) limits exists.
I never knew this, I think this would provide error except if we let t = 0 sooner, because:
\( \sum_{n=0}^{\infty}\frac{x^n}{n!} (\psi_0(n+1) - \ln(x)) = 0\\
(\sum_{n=0}^{\infty} \frac{x^n}{n!} \psi_0(n+1)) - \ln(x)e^x = 0\\
\)
the right series converges to \( \ln(x)e^x \) absolutely;
the left series' convergence depends on the digamma's function as it approaches infinity... Actually, writing this, I just thought of a neat proof.
\( \gamma = \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{k} - \ln(n) \)
and
\( \psi_0(n+1) = \sum_{k=1}^{n}\frac{1}{k} - \gamma \)
therefore:
\( L = \lim_{n\to\infty}|\frac{\frac{x^{n+1}}{n+1!}\psi_0(n+2)}{\frac{x^n}{n!} \psi_0(n+1)}| \)
\( L = \lim_{n\to\infty} |\frac{x}{n+1} \frac{\sum_{k=1}^{n+1} \frac{1}{k} - \gamma}{\sum_{c=1}^{n} \frac{1}{c} - \gamma}| \)
plug in our formula for the euler mascheroni constant
\( L = \lim_{n\to\infty} |\frac{x}{n+1} \frac{\sum_{k=1}^{n+1} \frac{1}{k} - \lim_{n\to\infty} \sum_{d=1}^{n} \frac{1}{d} - \ln(n)}{\sum_{c=1}^{n} \frac{1}{c} - \lim_{n\to\infty} \sum_{h=1}^{n} \frac{1}{h} - \ln(n)}| \)
\( L = \lim_{n\to\infty} |\frac{x}{n+1} \frac{\frac{1}{n+1} - \ln(n)}{-ln(n)}|\\
L = \lim_{n\to\infty} |\frac{x}{n+1} - \frac{x}{(n+1)^2\ln(n)}|\\
L = \lim_{n\to\infty} |x (\frac{1}{n+1} - \frac{1}{(n+1)^2\ln(n)})| \)
And now I'm stuck.... So I guess it comes to evaluating this to see if the series diverges. at least the first one, i'm still stumped as to why the second one doesn't converge if the first one does.
Is there a similar type of law for products of infinite series that I'm overlooking?
