What is the sum from n=1 to infinity of 2/(3n+8) ?

The sum from n=1 to infinity of 2/(3n+8) is diverges

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sum from n=1 to infinity of 2/(3n+8)

n = 1 \sum \infty \frac{2}{3 n + 8}

diverges

Solution steps

n = 1 \sum \infty \frac{2}{3 n + 8}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{3 n + 8}

Apply Series Limit Comparison Test:

diverges

= 2 diverges

= diverges

n = 1 \sum \infty \frac{1}{11 n ^{3} + 5}

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

n = 1 \sum \infty \frac{6}{n ^{2} + 6 n + 8}

n = 0 \sum \infty \frac{n}{5 n ^{3} + 1}

n = 0 \sum \infty 9^{\frac{1}{n}}

What is the sum from n=1 to infinity of 2/(3n+8) ?
The sum from n=1 to infinity of 2/(3n+8) is diverges