Proof of: ζ(s)=s−11k=1∑∞((k+1)sk−ksk−s)
Now,
ζ(s)=k=1∑∞ks1
ζ(s)=s∫1∞ts+1⌊t⌋dt
ζ(s)=s∫1∞ts+1t−{t}dt
ζ(s)=sk=1∑∞(∫kk+1ts+1t−t+kdt)
ζ(s)=sk=1∑∞(∫kk+1ts+1kdt)
ζ(s)=−11k=1∑∞((k+1)sk−(k)sk)
ζ(s)=s−11k=1∑∞((k+1)sk−(k)sk+(k)ss)
HENCE PROVED
ORIGINAL
#Calculus
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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
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I've proved it in the solution of this problem.