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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

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

Apply p-Series Test:

converges

= - converges

= converges

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

n = 0 \sum \infty \frac{n}{1 + n}

x = 0 \sum \infty \frac{x ^{x}}{x !}

n = 0 \sum \infty \frac{n}{n ^{2} + 3}

k = 1 \sum \infty \frac{4}{k ( k + 2 )}

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