Solution

For

ℜ s > 1

, absolute convergence permits separation into odd and even terms:

n = 1 \sum \infty (- 1)^{n - 1} n^{- s} = n = 1 \sum \infty n^{- s} - 2 n = 1 \sum \infty (2 n)^{- s} = (1 - 2^{1 - s}) ζ (s) .

Using the defining integral of the Gamma function and the substitution

u = n t

gives

Γ (s) n^{- s} = \int_{0}^{\infty} t^{s - 1} e^{- n t} d t .

Termwise integration is justified by absolute convergence when

ℜ s > 1

, and the geometric sum is

\sum_{n = 1}^{\infty} (- 1)^{n - 1} e^{- n t} = \frac{1}{1 + e ^{t}} .

Consequently the Dirichlet eta function satisfies

(1 - 2^{1 - s}) ζ (s) = η (s) = \frac{1}{Γ ( s )} \int_{0}^{\infty} \frac{t ^{s - 1}}{1 + e ^{t}} d t .

Near zero the integrand is

O (t^{ℜ s - 1})

, while at infinity it decays exponentially. The integral therefore defines a holomorphic function for

ℜ s > 0

. Thus

ζ (s) = \frac{1}{( 1 - 2 ^{1 - s} ) Γ ( s )} \int_{0}^{\infty} \frac{t ^{s - 1}}{1 + e ^{t}} d t

provides the desired continuation wherever the displayed denominator is nonzero. At a nonreal zero of

1 - 2^{1 - s}

, use instead

(1 - k^{1 - s}) ζ (s) = \sum_{n \geq 1} a_{n} n^{- s}, a_{n} = {1, 1 - k, k ∤ n, k ∣ n,

with an integer

k

for which

1 - k^{1 - s} \neq = 0

; the bounded partial sums of

a_{n}

give convergence for

ℜ s > 0

. This shows that those apparent singularities are removable and yields the Analytic continuation of the Riemann zeta function to the right half-plane.

s = 1

η (1) = \int_{0}^{\infty} \frac{d t}{1 + e ^{t}} = lo g 2,

whereas

1 - 2^{1 - s} = 1 - e^{- (s - 1) l o g 2} = (s - 1) lo g 2 + O ((s - 1)^{2}) .

Hence

ζ (s) = \frac{1}{s - 1} + O (1),

s = 1

is a simple pole and its residue is

Res_{s = 1} ζ (s) = 1 .

Solved by gpt-5.6-sol high.

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