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# Poof of Dirichlet series of Riemann zeta function

Proof of: $\zeta \left( s \right) =\frac { 1 }{ s-1 } \sum _{ k=1 }^{ \infty }{ \left( \frac { k }{ { \left( k+1 \right) }^{ s } } -\frac { k-s }{ { k }^{ s } } \right) }$

Now,

$$\displaystyle \zeta \left( s \right) =\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ s } } }$$

$$\displaystyle \zeta \left( s \right) = s \int _{ 1 }^{ \infty }{ \frac { \left\lfloor t \right\rfloor }{ { t }^{ s+1 } } dt }$$

$$\displaystyle \zeta \left( s \right) = s \int _{ 1 }^{ \infty }{ \frac { t-\left\{ t \right\} }{ { t }^{ s+1 } } dt }$$

$$\displaystyle \zeta \left( s \right) = s \sum _{ k=1 }^{ \infty }{ \left( \int _{ k }^{ k+1 }{ \frac { t-t+k }{ { t }^{ s+1 } } dt } \right) }$$

$$\displaystyle \zeta \left( s \right) =s\sum _{ k=1 }^{ \infty }{ \left( \int _{ k }^{ k+1 }{ \frac { k }{ { t }^{ s+1 } } dt } \right) }$$

$$\displaystyle \zeta \left( s \right) =-\frac { 1 }{ 1 } \sum _{ k=1 }^{ \infty }{ \left( \frac { k }{ { \left( k+1 \right) }^{ s } } -\frac { k }{ { \left( k \right) }^{ s } } \right) }$$

$$\displaystyle \zeta \left( s \right) =\frac { 1 }{ s-1 } \sum _{ k=1 }^{ \infty }{ \left( \frac { k }{ { \left( k+1 \right) }^{ s } } -\frac { k }{ { \left( k \right) }^{ s } } +\frac { s }{ { \left( k \right) }^{ s } } \right) }$$

$\large \text{HENCE PROVED}$

##### ORIGINAL

Note by Aditya Kumar
8 months, 1 week ago

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Nice.. If you elaborate more on how you get 2nd line from 1st line . it will help beginners.

I hope you will elaborate more · 8 months, 1 week ago

I've proved it in the solution of this problem. · 8 months, 1 week ago

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