View Full Version : Euler constants
zetafunction
Jul21-09, 04:22 AM
is the following sequence finite
\sum_{n=1}^{\infty} \frac{log^{u-1} (n)}{n} - u^{-1}log^{u}(n)
if u=1 then we have simply the Euler-Mascheroni constant but what happens in other cases or other values for 'u'
CRGreathouse
Jul21-09, 01:59 PM
I'm going to go out on a limb and guess that when you write
\sum_{n=1}^{\infty} \frac{log^{u-1} (n)}{n} - u^{-1}log^{u}(n)
you mean something like
\lim_{x\to\infty}-(\log x)^u+\sum_{n=1}^x\frac{(\log n)^{u-1}}{n}
but you may mean something else entirely.
zetafunction
Jul22-09, 07:57 AM
no but thanks by the answer i meant
\sum_{n=1}^{\infty} \frac{log^{u-1}(n)}{n} - \int_{1}^{\infty}\frac{log^{u-1}(x)}{x}
in case u=1 we have the Euler Mascheroni constant but how about for other values ??
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