What is the sum from n=0 to infinity of (-1^n)/(n+2) ?

The sum from n=0 to infinity of (-1^n)/(n+2) is diverges

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sum from n=0 to infinity of (-1^n)/(n+2)

n = 0 \sum \infty \frac{- 1 ^{n}}{n + 2}

diverges

Solution steps

n = 0 \sum \infty \frac{- 1 ^{n}}{n + 2}

Apply the constant multiplication rule:

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

= - n = 0 \sum \infty \frac{1 ^{n}}{n + 2}

Simplify

= - n = 0 \sum \infty \frac{1}{n + 2}

Apply Series Integral Test:

diverges

= - diverges

= diverges

\frac{d h}{d t} = - h^{2}

\int \frac{4}{x ^{4} - x ^{2}} d x

\int \frac{1}{x ^{2} x ^{2} - 5} d x

y = x^{8 c o s (x)}

y = arccsc (e^{9 t})

What is the sum from n=0 to infinity of (-1^n)/(n+2) ?
The sum from n=0 to infinity of (-1^n)/(n+2) is diverges