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

abitinomaq1

abitinomaq1

Answered question

2022-03-29

Proving identity
[(j+t)1j1]=k1ζ(k+1)(t)k

Answer & Explanation

Alejandra Hanna

Alejandra Hanna

Beginner2022-03-30Added 10 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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