What is the sum from n=1 to infinity}(-1)^n(\sqrt{n of)/(n+2) ?

The sum from n=1 to infinity}(-1)^n(\sqrt{n of)/(n+2) is converges

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sum from n=1 to infinity}(-1)^n(\sqrt{n of)/(n+2)

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

converges

Solution steps

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

Apply Alternating Series Test:

converges

= converges

\int \frac{289 - x ^{2}}{x} d x

x \to 0 lim ((1 + x)^{\frac{5}{x}})

x^{2} \frac{d y}{d x} - 2 x y = 6 y^{4}, y (1) = \frac{1}{2}

f (x) = x^{2} + 33, at x = 4

\frac{\partial}{\partial x} (2 x + 4 y^{2})

What is the sum from n=1 to infinity}(-1)^n(\sqrt{n of)/(n+2) ?
The sum from n=1 to infinity}(-1)^n(\sqrt{n of)/(n+2) is converges