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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

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

n = 1 \sum \infty (\frac{- e}{π})^{n}

n = 0 \sum \infty 3 (- 1)^{n}

n = 3 \sum \infty \frac{n ^{n}}{n !}

n = 3 \sum \infty \frac{4}{n ^{2} - 4}

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