What is the sum from n=1 to infinity of 5/(n+2) ?

The sum from n=1 to infinity of 5/(n+2) is diverges

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sum from n=1 to infinity of 5/(n+2)

n = 1 \sum \infty \frac{5}{n + 2}

diverges

Solution steps

n = 1 \sum \infty \frac{5}{n + 2}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{n + 2}

Apply Series Integral Test:

diverges

= 5 diverges

= diverges

\int (x^{3} + 3) d x

e^{x} \frac{d y}{d x} = e^{- y} + e^{2 x - y}

cos (x), sin (x)

2 x^{2} y^{^{''}} + 3 x y^{^{'}} = y

2 cos (7 x), 2 - 2 cos (7 x), 0 \leq x \leq \frac{π}{7}

What is the sum from n=1 to infinity of 5/(n+2) ?
The sum from n=1 to infinity of 5/(n+2) is diverges