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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

n = 1 \sum \infty (cos (9))^{n}

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

n = 1 \sum \infty n^{4}

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

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

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