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

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

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

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

1

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{2}

= 2 \cdot \frac{1}{2}

Simplify

2 \cdot \frac{1}{2} : 1

= 1

\int \frac{x ^{2} - 1}{x ^{3}} d x

\frac{d y}{d t} = 6 e^{2 t - y}

\frac{d}{d x} (m x e^{n x^{2}})

\frac{d}{d x} (5 \cdot T g (T, x) (2 x^{3}) + \frac{2}{x ^{2}})

y = \frac{5}{x}, (1, 5)

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