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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply p-Series Test:

converges

= - converges

= converges

n = 1 \sum \infty \frac{6}{3 n + 2}

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

n = 0 \sum \infty \frac{6}{n ( n + 2 )}

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

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

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