What is the sum from n=1 to infinity of 9/(sqrt(n)) ?

The sum from n=1 to infinity of 9/(sqrt(n)) is diverges

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sum from n=1 to infinity of 9/(sqrt(n))

n = 1 \sum \infty \frac{9}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{9}{n}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{1}{n}

Apply p-Series Test:

diverges

= 9 diverges

= diverges

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

n = 1 \sum \infty cos (\frac{π}{n})

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

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

n = 0 \sum \infty \frac{n}{( 1 + 4 n ) ^{2}}

What is the sum from n=1 to infinity of 9/(sqrt(n)) ?
The sum from n=1 to infinity of 9/(sqrt(n)) is diverges