What is the sum from n=1 to infinity of 2/(n+7) ?

The sum from n=1 to infinity of 2/(n+7) is diverges

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sum from n=1 to infinity of 2/(n+7)

n = 1 \sum \infty \frac{2}{n + 7}

diverges

Solution steps

n = 1 \sum \infty \frac{2}{n + 7}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{n + 7}

Apply Series Integral Test:

diverges

= 2 diverges

= diverges

a re a y = 6 x, y = 2 x^{2}

\frac{d}{d x} (sin^{3} (5 x^{2}))

\int \frac{1}{5 x ^{3}} d x

t an g e n t f (x) = x^{4} - 20 x^{2} + 64, at x = - 2

f a c t or 24 x^{3} + 24 x^{2}

What is the sum from n=1 to infinity of 2/(n+7) ?
The sum from n=1 to infinity of 2/(n+7) is diverges