What is the sum from n=1 to infinity of 8/(5n+12) ?

The sum from n=1 to infinity of 8/(5n+12) is diverges

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sum from n=1 to infinity of 8/(5n+12)

n = 1 \sum \infty \frac{8}{5 n + 12}

diverges

Solution steps

n = 1 \sum \infty \frac{8}{5 n + 12}

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty \frac{1}{5 n + 12}

Apply Series Integral Test:

diverges

= 8 diverges

= diverges

n = 0 \sum \infty \frac{2}{9 ^{n}}

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

n = 0 \sum \infty (- \frac{1}{8})^{n}

n = 1 \sum \infty 8 ln (\frac{n}{n + 1})

n = 0 \sum \infty 5 \cdot (\frac{1}{4})^{n}

What is the sum from n=1 to infinity of 8/(5n+12) ?
The sum from n=1 to infinity of 8/(5n+12) is diverges