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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

k = 2 \sum \infty (- 0.65)^{k}

n = 2 \sum \infty (\frac{3}{7})^{2 n}

n = 1 \sum \infty \frac{1}{1 - 4 n ^{2}}

n = 1 \sum \infty \frac{n ^{4}}{9 ^{n}}

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

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