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}

Simplify

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

Apply Series Integral Test:

diverges

= diverges

\int_{0}^{5} \frac{1}{( 6 - 2 x ) ^{\frac{1}{3}}} d x

\int (x^{3} - 2 x)^{2} (\frac{1}{x} - 5) d x

tan (e^{8 t}) + e^{t a n (8 t)}

f (x) = \frac{x}{e ^{x}}, at x = - 1

y^{^{'}} = 8 x e^{(5 y)}, y (0) = 0

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