What is the sum from n=3 to infinity of 1/(n^2-1) ?

The sum from n=3 to infinity of 1/(n^2-1) is converges

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sum from n=3 to infinity of 1/(n^2-1)

n = 3 \sum \infty \frac{1}{n ^{2} - 1}

converges

Solution steps

n = 3 \sum \infty \frac{1}{n ^{2} - 1}

Apply Series Integral Test:

converges

= converges

in v erse l a pl a ce \frac{s}{( s ^{2} + 9 ) ^{2}}

t an g e n t y = \frac{1}{4 - x}

\int \frac{( 4 x - 7 )}{x ^{2} - 6 x + 13} d x

d er i v a t i v e - \frac{7}{x ^{2}}

(x^{2} + 4) d y = (2 x - 8 x y) d x

What is the sum from n=3 to infinity of 1/(n^2-1) ?
The sum from n=3 to infinity of 1/(n^2-1) is converges