What is the sum from n=0 to infinity of n^{-2} ?

The sum from n=0 to infinity of n^{-2} is converges

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sum from n=0 to infinity of n^{-2}

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

converges

Solution steps

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

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

Apply p-Series Test:

converges

= converges

k = 3 \sum \infty 5 (\frac{1}{3})^{k}

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

n = 1 \sum \infty (\frac{5}{8})^{n - 1}

k = 2 \sum \infty (- 0.4)^{k}

n = 0 \sum \infty 7 e^{- 7 n}

What is the sum from n=0 to infinity of n^{-2} ?
The sum from n=0 to infinity of n^{-2} is converges