What is the sum from n=1 to infinity of 8/(sqrt(n)) ?

The sum from n=1 to infinity of 8/(sqrt(n)) is diverges

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sum from n=1 to infinity of 8/(sqrt(n))

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty \frac{1}{n}

Apply p-Series Test:

diverges

= 8 diverges

= diverges

n = 0 \sum \infty cos (\frac{1}{n ^{2}})

n = 1 \sum \infty \frac{1}{n ^{1.43}}

n = 1 \sum \infty \frac{9}{7 n ^{2} + 7 n}

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

n = 1 \sum \infty 9 ln (\frac{n}{n + 1})

What is the sum from n=1 to infinity of 8/(sqrt(n)) ?
The sum from n=1 to infinity of 8/(sqrt(n)) is diverges