Best solutions to - "How to prove∑n=p+1∞1n2−p2=12p(1+12+⋯+12p)"

Ekwadorkadba3

Answered question

2022-04-07

How to prove

\sum_{n = p + 1}^{\infty} \frac{1}{n^{2} - p^{2}} = \frac{1}{2 p} (1 + \frac{12}{+} \dots + \frac{1}{2 p})

davelucasnp0j

Beginner2022-04-08Added 16 answers

Use partial fractions to write

\sum_{n = p + 1}^{N} \frac{1}{n^{2} - p^{2}}

= \frac{1}{2 p} \sum_{n = p + 1}^{N} \frac{1}{n - p} - \frac{1}{n + p}

= \frac{1}{2 p} \sum_{n = 1}^{N - p} \frac{1}{n} - \frac{1}{2 p} \sum_{n = 2 p + 1}^{N + p} \frac{1}{n}

= \frac{1}{2 p} \sum_{n = 1}^{2 p} \frac{1}{n} - \frac{1}{2 p} \sum_{n = N - p + 1}^{N + p} \frac{1}{n}

Taking the limit of (1) as

N \to \infty

, we get

\sum_{n = p + 1}^{\infty} \frac{1}{n^{2} - p^{2}} = \frac{1}{2 p} \sum_{n = 1}^{2 p} \frac{1}{n}

ti2n1i2mt0l7

Beginner2022-04-09Added 10 answers

We have for

N \geq 2 p

\sum_{n = p + 1}^{N + p} \frac{1}{n^{2} - p^{2}} = \sum_{j = 1}^{N} \frac{1}{{(j + p)}^{2} - p^{2}}

= \sum_{j = 1}^{N} \frac{1}{j^{2} + 2 j p}

= \frac{1}{2 p} \sum_{j = 1}^{N} \frac{j + 2 p - j}{j (j + 2 p)}

= \frac{1}{2 p} (\sum_{j = 1}^{2 p} \frac{1}{j} + \sum_{j = 2 p + 1}^{N} \frac{1}{j} - \sum_{j = 2 p + 1}^{N + 2 p} \frac{1}{j})

= \frac{1}{2 p} (\sum_{j = 1}^{2 p} \frac{1}{j} - \sum_{j = N + 1}^{N + 2 p} \frac{1}{j})

\sum_{n = p + 1}^{N + p} \frac{1}{n^{2} - p^{2}} = \frac{1}{2 p} \sum_{j = 1}^{2 p} \frac{1}{j} - \frac{1}{2 p} \sum_{k = 1}^{2 p} \frac{1}{k + N}

We get the result taking the limit

N \to \infty

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