What is the sum from n=0 to infinity of 3/(sqrt(n)) ?

The sum from n=0 to infinity of 3/(sqrt(n)) is diverges

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sum from n=0 to infinity of 3/(sqrt(n))

n = 0 \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

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

n = 0 \sum \infty \frac{7}{1 0 ^{- n}}

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

n = 1 \sum \infty \frac{2}{3 n}

n = 1 \sum \infty \frac{2 n}{7 n - 4}

What is the sum from n=0 to infinity of 3/(sqrt(n)) ?
The sum from n=0 to infinity of 3/(sqrt(n)) is diverges