What is the sum from n=3 to infinity of 6/(n^2-1) ?

The sum from n=3 to infinity of 6/(n^2-1) is converges

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sum from n=3 to infinity of 6/(n^2-1)

n = 3 \sum \infty \frac{6}{n ^{2} - 1}

converges

Solution steps

n = 3 \sum \infty \frac{6}{n ^{2} - 1}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 6 \cdot n = 3 \sum \infty \frac{1}{n ^{2} - 1}

Apply Series Integral Test:

converges

= 6 converges

= converges

x^{5} + y^{5} = 10

x \to 0 - lim (\frac{1}{13 x})

f (x) = \frac{4 ( x ^{2} - 1 )}{( 4 x ^{3} - 2 x ) ^{\frac{2}{3}}}

\int t e^{- t^{2}} d t

x \to - \frac{π}{9} lim (9 x^{2} (5 x))

What is the sum from n=3 to infinity of 6/(n^2-1) ?
The sum from n=3 to infinity of 6/(n^2-1) is converges