What is the sum from n=0 to infinity of 2/(4n^2-1) ?

The sum from n=0 to infinity of 2/(4n^2-1) is -1

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sum from n=0 to infinity of 2/(4n^2-1)

n = 0 \sum \infty \frac{2}{4 n ^{2} - 1}

- 1

Solution steps

n = 0 \sum \infty \frac{2}{4 n ^{2} - 1}

Apply the constant multiplication rule:

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

= 2 \cdot n = 0 \sum \infty \frac{1}{4 n ^{2} - 1}

Apply Telescoping Series Test:

- \frac{1}{2}

= 2 (- \frac{1}{2})

Simplify

2 (- \frac{1}{2}) : - 1

= - 1

\int \frac{x + 2}{x ^{2} + 1} d x

y = 1 - x^{2} arcsin (x)

y^{^{'}} = (x^{2})

y^{^{'}} = \frac{( 5 x sec ( \frac{y}{x} ) + y )}{x}

\frac{d}{d x} (\frac{y}{1 - 2 x})

What is the sum from n=0 to infinity of 2/(4n^2-1) ?
The sum from n=0 to infinity of 2/(4n^2-1) is -1