What is the sum from n=1 to infinity of n^{-8} ?

The sum from n=1 to infinity of n^{-8} is converges

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sum from n=1 to infinity of n^{-8}

n = 1 \sum \infty n^{- 8}

converges

Solution steps

n = 1 \sum \infty n^{- 8}

Apply exponent rule:

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

= n = 1 \sum \infty \frac{1}{n ^{8}}

Apply p-Series Test:

converges

= converges

ln (\frac{q}{7})

\frac{\partial}{\partial x} (4 x e^{- x^{2} - y^{2} - z^{2}})

\int_{6}^{8} \frac{36}{( x - 6 ) ^{3}} d x

2 x^{2} ln (2 x)

\frac{\partial}{\partial x} (cos^{2} (2 x))

What is the sum from n=1 to infinity of n^{-8} ?
The sum from n=1 to infinity of n^{-8} is converges