What is the sum from n=1 to infinity of 3/(4n^2-1) ?

The sum from n=1 to infinity of 3/(4n^2-1) is 3/2

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sum from n=1 to infinity of 3/(4n^2-1)

n = 1 \sum \infty \frac{3}{4 n ^{2} - 1}

\frac{3}{2}

Solution steps

n = 1 \sum \infty \frac{3}{4 n ^{2} - 1}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{4 n ^{2} - 1}

Apply Telescoping Series Test:

\frac{1}{2}

= 3 \cdot \frac{1}{2}

Simplify

3 \cdot \frac{1}{2} : \frac{3}{2}

= \frac{3}{2}

1 + cos (2 x)

\frac{\partial}{\partial x} (\frac{100 x}{( x + y ) ^{2}})

\int 10 e^{- 2 x} d x

\frac{\partial}{\partial x} ((x^{2} + y^{2}) e^{x y})

\frac{d}{d x} (3 + 4 x)

What is the sum from n=1 to infinity of 3/(4n^2-1) ?
The sum from n=1 to infinity of 3/(4n^2-1) is 3/2