What is the sum from n=5 to infinity of (28)/(n+2) ?

The sum from n=5 to infinity of (28)/(n+2) is diverges

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sum from n=5 to infinity of (28)/(n+2)

n = 5 \sum \infty \frac{28}{n + 2}

diverges

Solution steps

n = 5 \sum \infty \frac{28}{n + 2}

Apply the constant multiplication rule:

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

= 28 \cdot n = 5 \sum \infty \frac{1}{n + 2}

Apply Series Integral Test:

diverges

= 28 diverges

= diverges

n = 2 \sum \infty \frac{n ^{3}}{n !}

n = 1 \sum \infty \frac{9 ^{n} + n}{n !}

n = 0 \sum \infty e^{5 n}

n = 1 \sum \infty cos (n + 1)

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

What is the sum from n=5 to infinity of (28)/(n+2) ?
The sum from n=5 to infinity of (28)/(n+2) is diverges