What is the sum from n=2 to infinity of 1/(n^{1.5)} ?

The sum from n=2 to infinity of 1/(n^{1.5)} is converges

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sum from n=2 to infinity of 1/(n^{1.5)}

n = 2 \sum \infty \frac{1}{n ^{1.5}}

converges

Solution steps

n = 2 \sum \infty \frac{1}{n ^{1.5}}

Apply exponent rule:

a^{n + m} = a^{n} \cdot a^{m}

n^{1.5} = n n

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

Apply Series Integral Test:

converges

= converges

y = 1 6^{s i n (x^{2})}

(2.3) (- 2.3)

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

\int \frac{x ^{5} - 9 x ^{3} - 9 x ^{2} - 9}{x ^{3} + 9 x} d x

\frac{d}{d x} (\frac{e ^{x}}{e ^{x} + 1})

What is the sum from n=2 to infinity of 1/(n^{1.5)} ?
The sum from n=2 to infinity of 1/(n^{1.5)} is converges