What is the sum from n=4 to infinity of 6/(n^2-1) ?

The sum from n=4 to infinity of 6/(n^2-1) is converges

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sum from n=4 to infinity of 6/(n^2-1)

n = 4 \sum \infty \frac{6}{n ^{2} - 1}

converges

Solution steps

n = 4 \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 = 4 \sum \infty \frac{1}{n ^{2} - 1}

Apply Series Integral Test:

converges

= 6 converges

= converges

x \to 0 lim (e^{- x (1 - s)})

x \to π lim (sin (x))

f (x) = 5 sin (x) - 3 cos (x), at x = π

\int 5 x e^{8 x} d x

\int \frac{e ^{x}}{e ^{x} + 2} d x

What is the sum from n=4 to infinity of 6/(n^2-1) ?
The sum from n=4 to infinity of 6/(n^2-1) is converges