What is the sum from n=1 to infinity of 3/(sqrt(n)) ?

The sum from n=1 to infinity of 3/(sqrt(n)) is diverges

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sum from n=1 to infinity of 3/(sqrt(n))

n = 1 \sum \infty \frac{3}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{3}{n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{n}

Apply p-Series Test:

diverges

= 3 diverges

= diverges

\int 9 4 - x^{2} d x

(\frac{1}{x ln ( x )})^{^{'}}

\frac{2 x ^{2} + 6 x + 4}{x}

\frac{d}{d x} (\frac{ln ( 3 x )}{7 x ^{2}})

f (x) = \frac{1}{x}

What is the sum from n=1 to infinity of 3/(sqrt(n)) ?
The sum from n=1 to infinity of 3/(sqrt(n)) is diverges