Poof of Dirichlet series of Riemann zeta function

Proof of: ζ(s)=1s−1∑k=1∞(k(k+1)s−k−sks)\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=1∞1ks\displaystyle \zeta \left( s \right) =\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ s } } }

ζ(s)=s∫1∞⌊t⌋ts+1dt\displaystyle \zeta \left( s \right) = s \int _{ 1 }^{ \infty }{ \frac { \left\lfloor t \right\rfloor }{ { t }^{ s+1 } } dt }

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

ζ(s)=s∑k=1∞(∫kk+1t−t+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)=s∑k=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)=−11∑k=1∞(k(k+1)s−k(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)=1s−1∑k=1∞(k(k+1)s−k(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
#Calculus

Note by Aditya Kumar
5 years, 4 months ago

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Comments

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 - 5 years, 4 months ago

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

Aditya Kumar - 5 years, 4 months ago
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