What is the sum from n=0 to infinity of (8n)/(6n+1) ?

The sum from n=0 to infinity of (8n)/(6n+1) is diverges

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sum from n=0 to infinity of (8n)/(6n+1)

n = 0 \sum \infty \frac{8 n}{6 n + 1}

diverges

Solution steps

n = 0 \sum \infty \frac{8 n}{6 n + 1}

Apply the constant multiplication rule:

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

= 8 \cdot n = 0 \sum \infty \frac{n}{6 n + 1}

Apply Series Divergence Test:

diverges

= 8 diverges

= diverges

x y^{^{''}} + 2 y^{^{'}} = 12 x^{2}, y^{^{'}} (1) = 4, y (1) = 0

(x^{2} + \frac{2 y}{x}) d x = (3 - ln (x^{2})) d y

\int 1 + 3 x d x

\int 28 x^{3} - 18 x^{2} + 10 x d x

\frac{d}{d x} (\frac{1}{x ^{\frac{2}{3}}})

What is the sum from n=0 to infinity of (8n)/(6n+1) ?
The sum from n=0 to infinity of (8n)/(6n+1) is diverges