What is the sum from n=1 to infinity of n^{-1/4} ?

The sum from n=1 to infinity of n^{-1/4} is diverges

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sum from n=1 to infinity of n^{-1/4}

n = 1 \sum \infty n^{- \frac{1}{4}}

diverges

Solution steps

n = 1 \sum \infty n^{- \frac{1}{4}}

Apply exponent rule:

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

= n = 1 \sum \infty \frac{1}{n ^{\frac{1}{4}}}

Apply p-Series Test:

diverges

= diverges

n = 0 \sum \infty arctan (\frac{1}{n})

n = 1 \sum \infty \frac{2 n + 1}{3 n}

n = 1 \sum \infty 3 (0.6)^{n - 1}

n = 1 \sum \infty \frac{2 ^{n}}{7 ^{n}}

n = 1 \sum \infty \frac{2}{6 ^{n}}

What is the sum from n=1 to infinity of n^{-1/4} ?
The sum from n=1 to infinity of n^{-1/4} is diverges