Explained: "Proving identity∑[(j+t)−1−j−1]=∑k≥1ζ(k+1)(−t)k"

abitinomaq1

Answered question

2022-03-29

Proving identity

\sum [{(j + t)}^{- 1} - j^{- 1}] = \sum_{k \geq 1} ζ (k + 1) {(- t)}^{k}

Alejandra Hanna

Beginner2022-03-30Added 10 answers

Just apply

\sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x}

. For

| t | < min j = 1

, we have

{(j + t)}^{- 1} = j^{- 1} {(1 + \frac{t}{j})}^{- 1}

= j^{- 1} \sum_{k = 0}^{\infty} {(- \frac{t}{j})}^{k}

Thus

\sum_{j = 1}^{\infty} \sum_{k = 1}^{\infty} \frac{{(- t)}^{k}}{j^{k + 1}} = \sum_{k = 1} \sum_{j = 1}^{\infty} \frac{{(- t)}^{k}}{j^{k + 1}}

= \sum_{k = 1}^{\infty} {(- t)}^{k} (\sum_{j = 1}^{\infty} \frac{1}{j^{k + 1}})

= \sum_{k = 1}^{\infty} {(- t)}^{k} ζ (k + 1)

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