$(1-2^{1-s})\zeta(s)$ is an entire function

Question

Show that $(1-2^{1-s})\zeta(s)$ is an entire function, which is represented by the series $$(1-2^{1-s})\zeta(s)=1-\dfrac{1}{2^s}+\dfrac{1}{3^s}-\dfrac{1}{4^s}+\cdots$$ for $\Re{s}>1$.

From the definition that $$\zeta(s)=\sum_{n=1}^\infty\dfrac{1}{n^s},$$ I want to multiply by $(1-2^{1-s})$. But it doesn't seem that the term $1-\dfrac{1}{2^s}+\ldots$ will come out.

Answer 1

Laurent series of $\zeta(s)$ about $1$ is $$\zeta(s) = \dfrac1{s-1} + \sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!} \gamma_k (s-1)^k$$ Hence, $$\eta(s) = \dfrac{1-2^{1-s}}{s-1} + \sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!} \gamma_k (s-1)^k(1-2^{1-s})$$ This makes sense everywhere except at $s=1$. At $s=1$, we have $$\lim_{s \to 1} \eta(s) = \lim_{s \to 1} \dfrac{1-2^{1-s}}{s-1} = \ln(2)$$ This shows the function is entire.

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For the series representation, see AaronS's answer.

Answer 2

For the series representation: \begin{align} \zeta(s) &= 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+ \cdots \\ 2^{1-s}\zeta(s) = \frac{2}{2^s}\zeta(s) &= 2(\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}+\frac{1}{8^s}+ \cdots) \end{align}

Subtract the two and the even terms will be negated: $$(1-2^{1-s})\zeta(s) = 1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+ \cdots$$

Answer 3

It seems that I am late to the party, but here is a conceptually clear proof that might be beneficial:

1. $ \zeta $ is meromorphic on $ \mathbb{C} $ with a pole of order $ 1 $ (a.k.a. a simple pole) at $ 1 $. (This happens to be the only non-trivial fact that we need, and we shall take it for granted.)

2. The mapping $ s \longmapsto 1 - 2^{1 - s} = 1 - e^{(1 - s) \ln(2)} $ is obviously entire.

3. The mapping $ \eta: s \longmapsto (1 - 2^{1 - s}) \zeta(s) $ is thus meromorphic on $ \mathbb{C} $ with either (i) a simple pole at $ 1 $ or (ii) a pole of order $ 0 $ (a.k.a. a removable singularity) at $ 1 $.

4. If (i) were the case, then $ \displaystyle \lim_{s \to 1^{+}} |\eta(s)| = \infty $.

5. Fortunately, this is not the case because $ \displaystyle \lim_{s \to 1^{+}} \eta(s) = \ln(2) $.

6. Therefore, Case (ii) holds.

7. Conclusion: $ \eta $ can be extended to an entire function whose value at $ 1 $ is $ \ln(2) $. The validity of the series representation has already been proven by Aaron.

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Note: In order to show that $ \displaystyle \lim_{s \to 1^{+}} \eta(s) = \lim_{s \to 1^{+}} (1 - 2^{1 - s}) \zeta(s) = \ln(2) $, we can employ the following formula (a college-level proof of which may be found here): $$ \forall s \in \mathbb{R}_{> 1}: \quad \zeta(s) = s \left( \frac{1}{s - 1} - \underbrace{\int_{1}^{\infty} \frac{\{ t \}}{t^{s + 1}} \mathrm{d}{t}} _{\text{Bounded above by $ 1 $}} \right), $$ where $ \{ \cdot \}: \mathbb{R} \to [0,1) $ is the fractional-part function. No more is needed than a simple application of l’Hôpital’s Rule to establish that $$ \lim_{s \to 1^{+}} \frac{1 - 2^{1 - s}}{s - 1} = \ln(2). $$