Poof of Dirichlet series of Riemann zeta function

Proof of: ζ(s)=1s1k=1(k(k+1)sksks)\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,

ζ(s)=k=11ks\displaystyle \zeta \left( s \right) =\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ s } } }

ζ(s)=s1tts+1dt\displaystyle \zeta \left( s \right) = s \int _{ 1 }^{ \infty }{ \frac { \left\lfloor t \right\rfloor }{ { t }^{ s+1 } } dt }

ζ(s)=s1t{t}ts+1dt\displaystyle \zeta \left( s \right) = s \int _{ 1 }^{ \infty }{ \frac { t-\left\{ t \right\} }{ { t }^{ s+1 } } dt }

ζ(s)=sk=1(kk+1tt+kts+1dt)\displaystyle \zeta \left( s \right) = s \sum _{ k=1 }^{ \infty }{ \left( \int _{ k }^{ k+1 }{ \frac { t-t+k }{ { t }^{ s+1 } } dt } \right) }

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

ζ(s)=11k=1(k(k+1)sk(k)s)\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) }

ζ(s)=1s1k=1(k(k+1)sk(k)s+s(k)s)\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) }

HENCE PROVED\large \text{HENCE PROVED}

ORIGINAL

Note by Aditya Kumar
3 years, 8 months 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

Aman Rajput - 3 years, 8 months ago

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I've proved it in the solution of this problem.

Aditya Kumar - 3 years, 8 months ago

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