What is the sum from n=0 to infinity of 6/(1-n^2) ?

The sum from n=0 to infinity of 6/(1-n^2) is -9/2

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sum from n=0 to infinity of 6/(1-n^2)

n = 0 \sum \infty \frac{6}{1 - n ^{2}}

- \frac{9}{2}

Solution steps

n = 2 \sum \infty \frac{6}{1 - n ^{2}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 2 \sum \infty \frac{1}{1 - n ^{2}}

Factor

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

= 6 \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}

= 6 (- 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 )}

= 6 (- n = 2 \sum \infty - \frac{1}{2 ( n + 1 )} + \frac{1}{2 ( n - 1 )})

Expand telescoping series:

\frac{3}{4}

Simplify

6 (- \frac{3}{4}) : - \frac{9}{2}

= - \frac{9}{2}

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

n = 1 \sum \infty 9^{- n}

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

k = 2 \sum \infty \frac{1}{e ^{k}}

n = 0 \sum \infty \frac{3 ^{n}}{e ^{n}}

What is the sum from n=0 to infinity of 6/(1-n^2) ?
The sum from n=0 to infinity of 6/(1-n^2) is -9/2