Convergence of Riemann's Zeta-function

Today I have played around with infinite series as well as with the famous Zeta function. And I did a litte proof that the zeta-function converges for sC s\in \mathbb{C} and Re(s)>2Re\left( s \right) > 2. I am new in this area of mathmatics and therefore I want to check (by contributing it) wether this kind of argumentation is valid.

The Zeta function is definded by: ζ(s)=11s+12s+13s+=n=11ns\zeta \left( s \right) =\frac { 1 }{ { 1 }^{ s } } +\frac { 1 }{ { 2 }^{ s } } +\frac { 1 }{ { 3 }^{ s } } +\dots =\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ s } } }

Let An=1n2{ A }_{ n }=\frac { 1 }{ { n }^{ 2 } } be the nn-th term of ζ(2)\zeta \left( 2 \right) : ζ(2)=112+122+132+=A1+A2+A3+=An=π26\zeta \left( 2 \right) =\frac { 1 }{ { 1 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 3 }^{ 2 } } +\dots ={ A }_{ 1 }+{ A }_{ 2 }+{ A }_{ 3 }+\dots =\sum { { A }_{ n } } =\frac { { \pi }^{ 2 } }{ 6 }

This convergence was proven by Euler! Let Bn=1nk{ B }_{ n }=\frac { 1 }{ { n }^{ k } } be the nn-th term of ζ(k)\zeta \left( k \right) , where kk is a real number greater than 22: ζ(k)=11k+12k+13k+=B1+B2+B3+=Bn\zeta \left( k \right) =\frac { 1 }{ { 1 }^{ k } } +\frac { 1 }{ { 2 }^{ k } } +\frac { 1 }{ { 3 }^{ k } } +\dots =B_{ 1 }+B_{ 2 }+B_{ 3 }+\dots =\sum { B_{ n } } For now we don't know wether this converges or not!

We know that Bn>0{ B }_{ n }>0 for all natural numbers nn. Therefore Bn\sum { B_{ n } } is bounded below at 00 and also non decreasing. For each natural number nn and each real number kk greater than 22 it is obviously true, that: nkn2{ n }^{ k } \ge { n }^{ 2 } And hence: 1nk1n2BnAn\begin{aligned}\frac { 1 }{ { n }^{ k } } &\le \frac { 1 }{ { n }^{ 2 } } \\B_{ n }&\le A_{ n }\end{aligned} Therefore Bn\sum { B_{ n } } is also bounded above at An\sum { A_{ n } } and because Bn\sum { B_{ n } } is bounded above, bounded below and is non-decrasing, it must converge.

ζ(s)\therefore \zeta \left( s \right) converges for all real numbers s2s\ge 2

Now we can start with complex values for ss...

Let s=a+ibs=a+i\cdot b be a complex value for the Zeta-function and let Cn=1na+ibC_{n}=\frac{1}{n^{a+i\cdot b}} be the nn-th term of the Zeta-function, where a>2a>2: ζ(s)=ζ(a+ib)=11a+ib+12a+ib+13a+ib+=C1+C2+C3+=Cn\zeta \left( s \right) =\zeta \left( a+i\cdot b \right) =\frac { 1 }{ { 1 }^{ a+i\cdot b } } +\frac { 1 }{ { 2 }^{ a+i\cdot b } } +\frac { 1 }{ { 3 }^{ a+i\cdot b } } +\dots =C_{ 1 }+C_{ 2 }+C_{ 3 }+\dots =\sum { C_{ n } } Now there's a little bit algebra: Cn=1na+ibCn=1na1nibCn=1nani(b)Cn=1naei(b)ln(n)Cn=1na\begin{aligned}C_{ n }&=\frac { 1 }{ { n }^{ a+i\cdot b } }\\C_{ n } &=\frac { 1 }{ { n }^{ a } } \cdot \frac { 1 }{ { n }^{ i\cdot b } } \\ C_{ n }&=\frac { 1 }{ { n }^{ a } } \cdot { n }^{ i\cdot \left( -b \right) }\\ C_{ n }&=\frac { 1 }{ { n }^{ a } } \cdot { e }^{ i\cdot \left( -b \right) \cdot ln\left( n \right) }\\ \left| C_{ n } \right| &=\frac { 1 }{ \left| { n }^{ a } \right| } \end{aligned} Because nn is a natural number we can simplify this a tiny bit: Cn=1na\left| C_{ n } \right| =\frac { 1 }{ { n }^{ a } }

There is a convergency test for complex infinite series, which states:

If An\sum{A_{n}} converges absolutely and CnAn\left| C_{n} \right| \le \left| A_{n} \right| for every natural number nn then Cn\sum{C_{n}} converges as well.

Because An=An=1n2\left| A_{n}\right|=A_{n}=\frac{1}{n^2}, An\sum{A_{n}} converges absolutely. Cn=1na1n2=An \left| C_{ n } \right| = \frac { 1 }{ { n }^{ a }} \le \frac{1}{n^2} = \left| A_{n}\right|

This inequality for real numbers a>2a >2 was proven in the first section of this note. Therefore Cn\sum{C_{n}} must converge. And finally:

ζ(s)\therefore \zeta \left( s \right) converges for all complex numbers ss where Re(s)>2Re\left( s \right) >2 .

I know that there is a little gap in the domain of the Zeta-function for which the convergence isn't proven by this note (all complex number ss where Re(s)(12]Re\left( s \right) \in \left( { 1 }|{ 2 } \right] ), but this note is just a result of a boring afternoon. I'm wondered wether you can prove this gap with the same kind of argumentation or wether you will need a new base for the complete proof. (I even don't know wether this proof is valid, that's the main reason why I contribute this note)

Note by CodeCrafter 1
1 week, 4 days ago

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