What is the sum from n=1 to infinity of 2/(n^{0.45)} ?

The sum from n=1 to infinity of 2/(n^{0.45)} is diverges

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sum from n=1 to infinity of 2/(n^{0.45)}

n = 1 \sum \infty \frac{2}{n ^{0.45}}

diverges

Solution steps

n = 1 \sum \infty \frac{2}{n ^{0.45}}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{n ^{0.45}}

Apply p-Series Test:

diverges

= 2 diverges

= diverges

n = 0 \sum \infty \frac{n}{10 n ^{6} + 1}

n = 0 \sum \infty \frac{3}{2 ^{n - 1}}

n = 1 \sum \infty \frac{5 ^{n}}{8}

n = 0 \sum \infty \frac{n}{49 + n ^{2}}

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

What is the sum from n=1 to infinity of 2/(n^{0.45)} ?
The sum from n=1 to infinity of 2/(n^{0.45)} is diverges