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Question Number 2702 by Filup last updated on 25/Nov/15
ζ(s)=Σ_(i=1) ^∞ i^(−s) =1+(1/2^s )+(1/3^s )+... Is ζ(s)>0∀s∈R? 1. Can you prove, or prove otherwise? 2. If ζ(s)>n, s∈R, what are the bounds of s? i.e. a≤s≤b : ζ(s)>n
$$\zeta\left({s}\right)=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{i}^{−{s}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{{s}} }+\frac{\mathrm{1}}{\mathrm{3}^{{s}} }+… \\ $$$$ \\ $$$$\mathrm{Is}\:\zeta\left({s}\right)>\mathrm{0}\forall{s}\in\mathbb{R}? \\ $$$$\mathrm{1}.\:\mathrm{Can}\:\mathrm{you}\:\mathrm{prove},\:\mathrm{or}\:\mathrm{prove}\:\mathrm{otherwise}? \\ $$$$\mathrm{2}.\:\mathrm{If}\:\zeta\left({s}\right)>{n},\:{s}\in\mathbb{R},\:\mathrm{what}\:\mathrm{are}\:\mathrm{the}\:\mathrm{bounds} \\ $$$$\mathrm{of}\:{s}?\:\mathrm{i}.\mathrm{e}.\:\:{a}\leqslant{s}\leqslant{b}\::\:\zeta\left({s}\right)>{n} \\ $$
Commented by Filup last updated on 25/Nov/15
i have actually solved that about a week ago. ill post it later I have work now i′ll make a new post later with the proof
$$\mathrm{i}\:\mathrm{have}\:\mathrm{actually}\:\mathrm{solved}\:\mathrm{that}\:\mathrm{ab}{o}\mathrm{ut}\:\mathrm{a}\:\mathrm{week} \\ $$$$\mathrm{ago}.\:\mathrm{ill}\:\mathrm{post}\:\mathrm{it}\:\mathrm{later} \\ $$$$\mathrm{I}\:{have}\:{work}\:\mathrm{now} \\ $$$$ \\ $$$$\mathrm{i}'\mathrm{ll}\:\mathrm{make}\:\mathrm{a}\:\mathrm{new}\:\mathrm{post}\:\mathrm{later}\:\mathrm{with}\:\mathrm{the}\:\mathrm{proof} \\ $$
Commented by Filup last updated on 25/Nov/15
Feel free to express ζ(s) in other froms of notation (such as in terms of Γ and integral notation)
$$\mathrm{Feel}\:\mathrm{free}\:\mathrm{to}\:\mathrm{express}\:\zeta\left({s}\right)\:\mathrm{in}\:\mathrm{other}\:\mathrm{froms} \\ $$$$\mathrm{of}\:\mathrm{notation}\:\left({such}\:{as}\:{in}\:{terms}\:{of}\right. \\ $$$$\left.\Gamma\:{and}\:{integral}\:{notation}\right) \\ $$
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 divergent.
$$\zeta\left({s}\right)\:\mathrm{converges}\:\mathrm{only}\:\mathrm{for}\:{s}>\mathrm{1}. \\ $$$$\mathrm{For}\:\mathrm{example} \\ $$$${s}=−\mathrm{2} \\ $$$$\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +..\:\mathrm{is}\:{divergent}. \\ $$$$ \\ $$
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+∙∙∙ wich diverge but S=1+2(1+2+∙∙∙) S=1+2S S=−1=(1/(1−2)) this is called analytic continuation its are crazy, but cool
$$\mathrm{however}\:\mathrm{by}\:\mathrm{some}\:\mathrm{ways}\:\mathrm{you}\:\mathrm{still}\:\mathrm{can}\:\mathrm{give} \\ $$$$\mathrm{a}\:\mathrm{value}\:\mathrm{to}\:\mathrm{it},\:\mathrm{like} \\ $$$$\mathrm{S}=\mathrm{1}+\mathrm{2}+\mathrm{4}+\centerdot\centerdot\centerdot\:\mathrm{wich}\:\mathrm{diverge}\:\mathrm{but} \\ $$$$\mathrm{S}=\mathrm{1}+\mathrm{2}\left(\mathrm{1}+\mathrm{2}+\centerdot\centerdot\centerdot\right) \\ $$$$\mathrm{S}=\mathrm{1}+\mathrm{2S} \\ $$$$\mathrm{S}=−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{called}\:\mathrm{analytic}\:\mathrm{continuation} \\ $$$$\mathrm{its}\:\mathrm{are}\:\mathrm{crazy},\:\mathrm{but}\:\mathrm{cool} \\ $$
Commented by prakash jain last updated on 25/Nov/15
Filup has posted some question earlier as well related to that. Σ_(i=1) ^∞ i=((−1)/(12)) or ζ(−1)=((−1)/(12)) I will read more to see what are the possibilities for ζ function, for example ζ(−2)=Σ_(i=1) ^∞ i^2
$$\mathrm{Filup}\:\mathrm{has}\:\mathrm{posted}\:\mathrm{some}\:\mathrm{question}\:\mathrm{earlier}\:\mathrm{as} \\ $$$$\mathrm{well}\:\mathrm{related}\:\mathrm{to}\:\mathrm{that}. \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{i}=\frac{−\mathrm{1}}{\mathrm{12}}\:\mathrm{or}\:\zeta\left(−\mathrm{1}\right)=\frac{−\mathrm{1}}{\mathrm{12}} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{read}\:\mathrm{more}\:\mathrm{to}\:\mathrm{see}\:\mathrm{what}\:\mathrm{are}\:\mathrm{the}\:\mathrm{possibilities} \\ $$$$\mathrm{for}\:\zeta\:\mathrm{function},\:\mathrm{for}\:\mathrm{example}\:\zeta\left(−\mathrm{2}\right)=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{i}^{\mathrm{2}} \\ $$
Commented by 123456 last updated on 25/Nov/15
there a functional equation to achive it by this you get ζ(−2n)=0 n∈N^∗ the functional equation is ζ(s)=2^s π^(s−1) sin (((πs)/2))Γ(1−s)ζ(1−s) its crazy, but cool
$$\mathrm{there}\:\mathrm{a}\:\mathrm{functional}\:\mathrm{equation}\:\mathrm{to}\:\mathrm{achive}\:\mathrm{it} \\ $$$$\mathrm{by}\:\mathrm{this}\:\mathrm{you}\:\mathrm{get} \\ $$$$\zeta\left(−\mathrm{2}{n}\right)=\mathrm{0}\:\:\:{n}\in\mathbb{N}^{\ast} \\ $$$$\mathrm{the}\:\mathrm{functional}\:\mathrm{equation}\:\mathrm{is} \\ $$$$\zeta\left({s}\right)=\mathrm{2}^{{s}} \pi^{{s}−\mathrm{1}} \mathrm{sin}\:\left(\frac{\pi{s}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−{s}\right)\zeta\left(\mathrm{1}−{s}\right) \\ $$$$\mathrm{its}\:\mathrm{crazy},\:\mathrm{but}\:\mathrm{cool} \\ $$
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(−2n)=0, that′d be awsome
$$\mathrm{I}\:\mathrm{was}\:\mathrm{incorrect}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{not}\:\mathrm{done}\:\mathrm{the} \\ $$$$\mathrm{proof}\:\mathrm{I}\:\mathrm{said}\:\mathrm{I}\:\mathrm{did}.\:\mathrm{I}\:\mathrm{mistakenly}\:\mathrm{mixed}\: \\ $$$$\mathrm{it}\:\mathrm{up}\:\mathrm{with}\:\mathrm{something}\:\mathrm{else}.\:\mathrm{If}\:\mathrm{someone}\:\mathrm{could} \\ $$$$\mathrm{make}\:\mathrm{a}\:\mathrm{post}\:\mathrm{on}\:\mathrm{how}\:\zeta\left(−\mathrm{2}{n}\right)=\mathrm{0},\:\mathrm{that}'\mathrm{d} \\ $$$$\mathrm{be}\:\mathrm{awsome} \\ $$
Commented by 123456 last updated on 26/Nov/15
in general ζ(s)≥1,s>1,s∈R later i add the answer its follow a similiar ideia than proof that lim_(n→+∞) H_n −ln n=γ
$$\mathrm{in}\:\mathrm{general} \\ $$$$\zeta\left({s}\right)\geqslant\mathrm{1},{s}>\mathrm{1},{s}\in\mathbb{R} \\ $$$$\mathrm{later}\:\mathrm{i}\:\mathrm{add}\:\mathrm{the}\:\mathrm{answer} \\ $$$$\mathrm{its}\:\mathrm{follow}\:\mathrm{a}\:\mathrm{similiar}\:\mathrm{ideia}\:\mathrm{than} \\ $$$$\mathrm{proof}\:\mathrm{that}\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\mathrm{H}_{{n}} −\mathrm{ln}\:{n}=\gamma \\ $$
Answered by 123456 last updated on 27/Nov/15
x≥1,s>1 ∫_1 ^(+∞) (dx/(⌈x⌉^s ))=Σ_(n=1) ^(+∞) (1/((n+1)^s )) ∫_1 ^(+∞) (dx/(⌊x⌋^s ))=Σ_(n=1) ^(+∞) (1/n^s ) 0<(1/(⌈x⌉^s ))≤(1/x^s )≤(1/(⌊x⌋^s )) (= only for x∈N^∗ ) 0<∫_1 ^(+∞) (dx/(⌈x⌉^s ))<∫_1 ^(+∞) (dx/x^s )<∫_1 ^(+∞) (dx/(⌊x⌋^s )) 0<Σ_(n=1) ^(+∞) (1/((n+1)^s ))<(1/(s−1))<Σ_(n=1) ^(+∞) (1/n^s ) 1<1+Σ_(n=1) ^(+∞) (1/((n+1)^s ))<1+(1/(s−1)) 1<Σ_(n=1) ^(+∞) (1/n^s )<(s/(s−1)) so 1<(1/(s−1))<ζ(s)<(s/(s−1)) (s>1) s→+∞,ζ(s)→1 s→1^+ ,ζ(s)→+∞
$${x}\geqslant\mathrm{1},{s}>\mathrm{1} \\ $$$$\underset{\mathrm{1}} {\overset{+\infty} {\int}}\frac{{dx}}{\lceil{x}\rceil^{{s}} }=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{s}} } \\ $$$$\underset{\mathrm{1}} {\overset{+\infty} {\int}}\frac{{dx}}{\lfloor{x}\rfloor^{{s}} }=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{s}} } \\ $$$$\mathrm{0}<\frac{\mathrm{1}}{\lceil{x}\rceil^{{s}} }\leqslant\frac{\mathrm{1}}{{x}^{{s}} }\leqslant\frac{\mathrm{1}}{\lfloor{x}\rfloor^{{s}} }\:\:\left(=\:\mathrm{only}\:\mathrm{for}\:{x}\in\mathbb{N}^{\ast} \right) \\ $$$$\mathrm{0}<\underset{\mathrm{1}} {\overset{+\infty} {\int}}\frac{{dx}}{\lceil{x}\rceil^{{s}} }<\underset{\mathrm{1}} {\overset{+\infty} {\int}}\frac{{dx}}{{x}^{{s}} }<\underset{\mathrm{1}} {\overset{+\infty} {\int}}\frac{{dx}}{\lfloor{x}\rfloor^{{s}} } \\ $$$$\mathrm{0}<\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{s}} }<\frac{\mathrm{1}}{{s}−\mathrm{1}}<\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{s}} } \\ $$$$\mathrm{1}<\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{s}} }<\mathrm{1}+\frac{\mathrm{1}}{{s}−\mathrm{1}} \\ $$$$\mathrm{1}<\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{s}} }<\frac{{s}}{{s}−\mathrm{1}} \\ $$$$\mathrm{so} \\ $$$$\mathrm{1}<\frac{\mathrm{1}}{{s}−\mathrm{1}}<\zeta\left({s}\right)<\frac{{s}}{{s}−\mathrm{1}}\:\:\:\:\left({s}>\mathrm{1}\right) \\ $$$${s}\rightarrow+\infty,\zeta\left({s}\right)\rightarrow\mathrm{1} \\ $$$${s}\rightarrow\mathrm{1}^{+} ,\zeta\left({s}\right)\rightarrow+\infty \\ $$

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