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Is it true that for complex $s$,

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

If so, prove it.

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I changed the tag from [number-theory] to [complex-analysis] and [limit]. If this is homework then that's also a tag you should edit in yourself. –  anon Sep 22 '11 at 20:39
    
This is a piece of a problem from a Number Thoery book Im working through. I dont know much Complex Analysis. Any suggested books? –  Jason Smith Sep 22 '11 at 21:53
    
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. –  anon Sep 22 '11 at 21:57
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1 Answer

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 $\frac{2-2^s}{s-1}.$

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Since the limit we're trying to calculate is, by definition, the derivative of $-2^s$ at 1, it seems silly to use a method that requires you already to know this derivative. –  Chris Eagle Sep 22 '11 at 20:40
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Dear Eric, To amplify @Chris's point, one can see that this is (by definition) computing the derivative of $-2^s$ at $s = 1$, and hence compute it directly. There is no need to mention L'Hopital. Regards, –  Matt E Sep 22 '11 at 20:58
    
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. –  robjohn Sep 22 '11 at 21:38
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