What is the sum from n=1 to infinity of 2/(n+2) ?

The sum from n=1 to infinity of 2/(n+2) is diverges

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sum from n=1 to infinity of 2/(n+2)

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

diverges

= 2 diverges

= diverges

n = 1 \sum \infty \frac{5}{3 n ^{2}}

n = 1 \sum \infty \frac{1}{2 \cdot n - 1}

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

n = 1 \sum \infty e^{- \frac{n}{9}}

n = 1 \sum \infty \frac{1}{n ^{10}}

What is the sum from n=1 to infinity of 2/(n+2) ?
The sum from n=1 to infinity of 2/(n+2) is diverges