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

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

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sum from n=1 to infinity of (-1)^nn^{-2}

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

converges

Solution steps

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

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

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

Apply Alternating Series Test:

converges

= converges

\frac{1}{( s + 1 ) ^{2} ( s ^{2} + 1 )}

(x 4 - x^{2})^{^{'}}

n = 1 \sum \infty \frac{1 ^{n}}{n ^{7}}

x \to 0 lim (\frac{x ^{2}}{∣ x ∣})

y^{^{'}} = t^{2} y^{2}

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