What is the sum from n=1 to infinity of 4/(n^{-2.3)} ?

The sum from n=1 to infinity of 4/(n^{-2.3)} is diverges

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sum from n=1 to infinity of 4/(n^{-2.3)}

n = 1 \sum \infty \frac{4}{n ^{- 2.3}}

diverges

Solution steps

n = 1 \sum \infty \frac{4}{n ^{- 2.3}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 4 \cdot n = 1 \sum \infty \frac{1}{n ^{- 2.3}}

Apply Series Divergence Test:

diverges

= 4 diverges

= diverges

n = 0 \sum \infty (- 0.2)^{n}

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

n = 0 \sum \infty cos (\frac{3}{n ^{2}})

n = 2 \sum \infty \frac{1}{( n ln ( n ) )}

n = 1 \sum \infty \frac{4}{n ^{2} + n}

What is the sum from n=1 to infinity of 4/(n^{-2.3)} ?
The sum from n=1 to infinity of 4/(n^{-2.3)} is diverges