What is the sum from n=0 to infinity of (3n)/(6+n^2) ?

The sum from n=0 to infinity of (3n)/(6+n^2) is diverges

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sum from n=0 to infinity of (3n)/(6+n^2)

n = 0 \sum \infty \frac{3 n}{6 + n ^{2}}

diverges

Solution steps

n = 0 \sum \infty \frac{3 n}{6 + n ^{2}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty \frac{n}{6 + n ^{2}}

Apply Series Integral Test:

diverges

= 3 diverges

= diverges

n = 1 \sum \infty \frac{4}{3 ^{n} + 8}

k = 1 \sum \infty (cos (17))^{k}

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

n = 1 \sum \infty cos (\frac{15}{n})

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

What is the sum from n=0 to infinity of (3n)/(6+n^2) ?
The sum from n=0 to infinity of (3n)/(6+n^2) is diverges