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

The sum from n=0 to infinity of 1/(1-n^2) is -3/4

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

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

- \frac{3}{4}

Decimal

- 0.75

Solution steps

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

Factor

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

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

Apply the constant multiplication rule:

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

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

Take the partial fraction of

\frac{1}{- 1 + n ^{2}} : - \frac{1}{2 ( n + 1 )} + \frac{1}{2 ( n - 1 )}

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

Expand telescoping series:

\frac{3}{4}

= - \frac{3}{4}

What is the sum from n=0 to infinity of 1/(1-n^2) ?
The sum from n=0 to infinity of 1/(1-n^2) is -3/4