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

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

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

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

- \frac{1}{2}

Solution steps

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

Factor

1 - 4 n^{2} : - (- 1 + 4 n^{2})

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

Apply the constant multiplication rule:

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

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

Apply Telescoping Series Test:

\frac{1}{2}

= - \frac{1}{2}

n = 1 \sum \infty \frac{n ^{4}}{9 ^{n}}

n = 1 \sum \infty \frac{2 + sin ( n )}{n}

n = 2 \sum \infty (\frac{2}{7})^{2 n}

n = 1 \sum \infty 2 (0.7)^{n - 1}

n = 0 \sum \infty cos (n) π

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