n-1-1-n-2-1-

Question Number 114023 by Khalmohmmad last updated on 16/Sep/20

\underset{n = 1}{\sum^{\propto}} \frac{1}{n^{2} - 1} = ?

Commented by Dwaipayan Shikari last updated on 16/Sep/20

D i v e r g e s

f i r s t t e r m \frac{1}{n^{2} - 1} \to \infty (n = 1) o r u n d e f i n e d

Answered by mr W last updated on 16/Sep/20

\underset{n = 2}{\sum^{\propto}} \frac{1}{n^{2} - 1}

= \frac{1}{2} \underset{n = 2}{\sum^{\propto}} (\frac{1}{n - 1} - \frac{1}{n + 1})

= \frac{1}{2} (\frac{1}{1} - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \dots)

= \frac{1}{2} \times \frac{3}{2}

= \frac{3}{4}

Commented by Olaf last updated on 16/Sep/20

Erratum!

S_{n} = \frac{1}{2} (\underset{k = 2}{\sum^{n}} \frac{1}{k - 1} - \underset{k = 2}{\sum^{n}} \frac{1}{k + 1})

S_{n} = \frac{1}{2} (\underset{k = 1}{\sum^{n - 1}} \frac{1}{k} - \underset{k = 3}{\sum^{n + 1}} \frac{1}{k})

S_{n} = \frac{1}{2} (1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n + 1})

Double subscripts: use braces to clarify

Commented by mr W last updated on 16/Sep/20

y o u a r e r i g h t!

Answered by Olaf last updated on 16/Sep/20

for n = 1, \frac{1}{n^{2} - 1} is not defined!

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