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Question Number 2702 by Filup last updated on 25/Nov/15

ζ (s) = \underset{i = 1}{\sum^{\infty}} i^{- s} = 1 + \frac{1}{2^{s}} + \frac{1}{3^{s}} + \dots

Is ζ (s) > 0 \forall s \in R ?

1 . Can you prove, or prove otherwise ?

2 . If ζ (s) > n, s \in R, what are the bounds

of s ? i . e . a ⩽ s ⩽ b : ζ (s) > n

Commented by Filup last updated on 25/Nov/15

i have actually solved that ab o ut a week

ago . ill post it later

I h a v e w o r k now

i^{'} ll make a new post later with the proof

Commented by Filup last updated on 25/Nov/15

Feel free to express ζ (s) in other froms

of notation (s u c h a s i n t e r m s o f

Γ a n d i n t e g r a l n o t a t i o n)

Commented by prakash jain last updated on 25/Nov/15

ζ (s) converges only for s > 1 .

For example

s = - 2

1 + 2^{2} + 3^{2} + . . is d i v e r g e n t .

Commented by 123456 last updated on 25/Nov/15

however by some ways you still can give

a value to it, like

S = 1 + 2 + 4 + \cdot \cdot \cdot wich diverge but

S = 1 + 2 (1 + 2 + \cdot \cdot \cdot)

S = 1 + 2 S

S = - 1 = \frac{1}{1 - 2}

this is called analytic continuation

its are crazy, but cool

Commented by prakash jain last updated on 25/Nov/15

Filup has posted some question earlier as

well related to that .

\underset{i = 1}{\sum^{\infty}} i = \frac{- 1}{12} or ζ (- 1) = \frac{- 1}{12}

I will read more to see what are the possibilities

for ζ function, for example ζ (- 2) = \underset{i = 1}{\sum^{\infty}} i^{2}

Commented by 123456 last updated on 25/Nov/15

there a functional equation to achive it

by this you get

ζ (- 2 n) = 0 n \in N^{*}

the functional equation is

ζ (s) = 2^{s} π^{s - 1} \sin (\frac{π s}{2}) Γ (1 - s) ζ (1 - s)

its crazy, but cool

Commented by Filup last updated on 26/Nov/15

I was incorrect . I have not done the

proof I said I did . I mistakenly mixed

it up with something else . If someone could

make a post on how ζ (- 2 n) = 0, {that}^{'} d

be awsome

Commented by 123456 last updated on 26/Nov/15

in general

ζ (s) ⩾ 1, s > 1, s \in R

later i add the answer

its follow a similiar ideia than

proof that \lim_{n \to + \infty} H_{n} - \ln n = γ

Answered by 123456 last updated on 27/Nov/15

x ⩾ 1, s > 1

\underset{1}{\int^{+ \infty}} \frac{d x}{⌈ x ⌉^{s}} = \underset{n = 1}{\sum^{+ \infty}} \frac{1}{{(n + 1)}^{s}}

\underset{1}{\int^{+ \infty}} \frac{d x}{⌊ x ⌋^{s}} = \underset{n = 1}{\sum^{+ \infty}} \frac{1}{n^{s}}

0 < \frac{1}{⌈ x ⌉^{s}} ⩽ \frac{1}{x^{s}} ⩽ \frac{1}{⌊ x ⌋^{s}} (= only for x \in N^{*})

0 < \underset{1}{\int^{+ \infty}} \frac{d x}{⌈ x ⌉^{s}} < \underset{1}{\int^{+ \infty}} \frac{d x}{x^{s}} < \underset{1}{\int^{+ \infty}} \frac{d x}{⌊ x ⌋^{s}}

0 < \underset{n = 1}{\sum^{+ \infty}} \frac{1}{{(n + 1)}^{s}} < \frac{1}{s - 1} < \underset{n = 1}{\sum^{+ \infty}} \frac{1}{n^{s}}

1 < 1 + \underset{n = 1}{\sum^{+ \infty}} \frac{1}{{(n + 1)}^{s}} < 1 + \frac{1}{s - 1}

1 < \underset{n = 1}{\sum^{+ \infty}} \frac{1}{n^{s}} < \frac{s}{s - 1}

so

1 < \frac{1}{s - 1} < ζ (s) < \frac{s}{s - 1} (s > 1)

s \to + \infty, ζ (s) \to 1

s \to 1^{+}, ζ (s) \to + \infty

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