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

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

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

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

2

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Telescoping Series Test:

\frac{1}{2}

= 4 \cdot \frac{1}{2}

Simplify

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

= 2

\frac{\partial}{\partial x} (x^{2} y z^{2} - 4 cos (3 yz))

f (x) = - 3 x^{2} + 1 x, at x = 0

\frac{d}{d x} (1 + 3 sin^{3} (x))

x^{2} + 1 + x^{2}

\int_{\frac{π}{4}}^{x} sec^{2} (t) d t

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