Proving identity \sum[(j+t)^{-1}-j^{-1}]=\sum_{k\geq1}\zeta(k+1)(-t)^k

Dashawn Robbins 2022-05-01 Answered
Proving identity
[(j+t)1j1]=k1ζ(k+1)(t)k
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Answers (1)

jeffster830gyz
Answered 2022-05-02 Author has 21 answers
Just apply n=0xn=11x. For |t|<minj=1, we have
(j+t)1=j1(1+tj)1
=j1k=0(tj)k
Thus
j=1k=1(t)kjk+1=k=1j=1(t)kjk+1
=k=1(t)k(j=11jk+1)
=k=1(t)kζ(k+1)
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