What is the sum from n=1 to infinity of (3n)/(6n+13) ?

The sum from n=1 to infinity of (3n)/(6n+13) is diverges

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sum from n=1 to infinity of (3n)/(6n+13)

n = 1 \sum \infty \frac{3 n}{6 n + 13}

diverges

Solution steps

n = 1 \sum \infty \frac{3 n}{6 n + 13}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{n}{6 n + 13}

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

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What is the sum from n=1 to infinity of (3n)/(6n+13) ?
The sum from n=1 to infinity of (3n)/(6n+13) is diverges