What is the sum from n=1 to infinity of 1/(n^{-6)} ?

The sum from n=1 to infinity of 1/(n^{-6)} is diverges

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sum from n=1 to infinity of 1/(n^{-6)}

n = 1 \sum \infty \frac{1}{n ^{- 6}}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{- 6}}

Simplify

\frac{1}{n ^{- 6}} : n^{6}

= n = 1 \sum \infty n^{6}

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty ln (\frac{n + 3}{n})

n = 1 \sum \infty - \frac{1}{1 3 ^{n}}

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

j = 1 \sum \infty 5^{- 3 j}

n = 0 \sum \infty \frac{1}{n ^{8}}

What is the sum from n=1 to infinity of 1/(n^{-6)} ?
The sum from n=1 to infinity of 1/(n^{-6)} is diverges