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

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

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

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

- \frac{3}{2}

Solution steps

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

Apply the constant multiplication rule:

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

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

Factor

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

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

Apply the constant multiplication rule:

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

= 2 (- 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 )}

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

Expand telescoping series:

\frac{3}{4}

Simplify

2 (- \frac{3}{4}) : - \frac{3}{2}

= - \frac{3}{2}

n = 1 \sum \infty 5 \cdot 0. 2^{n - 1}

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

n = 1 \sum \infty (0.6)^{n - 1}

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

n = 0 \sum \infty \frac{1}{n ^{0.9}}

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