# 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 $$s\in \mathbb{C}$$ and $$Re\left( s \right) > 2$$. 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: $\zeta \left( s \right) =\frac { 1 }{ { 1 }^{ s } } +\frac { 1 }{ { 2 }^{ s } } +\frac { 1 }{ { 3 }^{ s } } +\dots =\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ s } } }$

Let ${ A }_{ n }=\frac { 1 }{ { n }^{ 2 } }$ be the $n$-th term of $\zeta \left( 2 \right)$: $\zeta \left( 2 \right) =\frac { 1 }{ { 1 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 3 }^{ 2 } } +\dots ={ A }_{ 1 }+{ A }_{ 2 }+{ A }_{ 3 }+\dots =\sum { { A }_{ n } } =\frac { { \pi }^{ 2 } }{ 6 }$

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

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

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

Now we can start with complex values for $s$...

Let $s=a+i\cdot b$ be a complex value for the Zeta-function and let $C_{n}=\frac{1}{n^{a+i\cdot b}}$ be the $n$-th term of the Zeta-function, where $a>2$: $\zeta \left( s \right) =\zeta \left( a+i\cdot b \right) =\frac { 1 }{ { 1 }^{ a+i\cdot b } } +\frac { 1 }{ { 2 }^{ a+i\cdot b } } +\frac { 1 }{ { 3 }^{ a+i\cdot b } } +\dots =C_{ 1 }+C_{ 2 }+C_{ 3 }+\dots =\sum { C_{ n } }$ Now there's a little bit algebra: \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 $n$ is a natural number we can simplify this a tiny bit: $\left| C_{ n } \right| =\frac { 1 }{ { n }^{ a } }$

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

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

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

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

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

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 $s$ where $Re\left( s \right) \in \left( { 1 }|{ 2 } \right]$), 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 year, 3 months ago

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