What is the sum from n=0 to infinity of 6/(sqrt(n)) ?

The sum from n=0 to infinity of 6/(sqrt(n)) is diverges

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sum from n=0 to infinity of 6/(sqrt(n))

n = 0 \sum \infty \frac{6}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{6}{n}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{n}

Apply p-Series Test:

diverges

= 6 diverges

= diverges

i = 1 \sum \infty e^{- i}

n = 1 \sum \infty \frac{( - 3 + 2 ) ^{n}}{n}

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

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

n = 0 \sum \infty 8 + 4 + 2 + 1

What is the sum from n=0 to infinity of 6/(sqrt(n)) ?
The sum from n=0 to infinity of 6/(sqrt(n)) is diverges