What is the sum from n=0 to infinity of n^{-0.48} ?

The sum from n=0 to infinity of n^{-0.48} is diverges

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sum from n=0 to infinity of n^{-0.48}

n = 0 \sum \infty n^{- 0.48}

diverges

Solution steps

n = 1 \sum \infty n^{- 0.48}

Apply Series Integral Test:

diverges

= diverges

x \to 1 lim ([f (x) + g (x)])

\frac{d}{d x} ((10 x)^{l n (10 x)})

\frac{\partial}{\partial x} (tan (x - y))

\int ln (1) d x

(x^{3} + e^{y}) d y = 3 x^{2} d x

What is the sum from n=0 to infinity of n^{-0.48} ?
The sum from n=0 to infinity of n^{-0.48} is diverges