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

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

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

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

converges

Solution steps

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

Simplify

\frac{1}{( 4 n ) ^{2} - 1} : \frac{1}{16 n ^{2} - 1}

= n = 1 \sum \infty \frac{1}{16 n ^{2} - 1}

Apply Series Integral Test:

converges

= converges

x \to - 1 + lim (- 1)

\frac{d}{d x} (x e^{4 x + 1} + \frac{1}{2})

f (x) = x^{2} - sec (x)

x y^{^{''}} = y^{^{'}}

s (t) = \frac{400 t}{( t ^{2 + 5} ) ^{3}}

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