2
$\begingroup$

Is it true that for complex $s$,

$\lim_{s\to 1} \frac{2-2^s}{s-1}=-2\log (2)?$

If so, prove it.

  • 0
    Dunno, I've only picked it up here and there. It's possible you could get the fundamentals down through lecture notes and pdfs and Wikipedia and such. Number theory (of the not-elementary sort), especially analytic NT, utilizes complex analysis a lot.2011-09-22

1 Answers 1

4

Yes it is true.

Added: The nicest solution is to just note that the above is the definition of the derivative.

Hint: Use l'Hopitals rule. Make sure you can justify why this is allowed now that we are dealing with complex numbers.

Alternative Hint: Use power series. Since $2^s$ is analytic everywhere, we can expand around $s=1$ and write it as $2e^{(s-1)\log 2}=2+2(s-1)\log 2+2\frac{(s-1)^2}{2!}\log^2 2+\cdots $ From this you can deduce the Laurent series around $s=1$ for $\frac{2-2^s}{s-1}.$

  • 0
    The factor of $2$ seems to have disappeared from all but the lead term of your series. I would have answered as in your Alternate Hint. However, for analytic functions, l'Hopital's rule is indeed valid.2011-09-22