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 diverges

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

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

diverges

Solution steps

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

Simplify

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

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

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{1}{9 - n}

n = 0 \sum \infty \frac{1}{1 - 7 n}

n = 0 \sum \infty \frac{1}{8 n + 4}

n = 0 \sum \infty \frac{n !}{9 ^{n}}

n = 1 \sum \infty \frac{9}{n ^{2} + 16}

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 diverges