What is the sum from n=1 to infinity}((-1)^n(2)^{n-1})/((2)^{n+1)(n)^{1/3 of} ?

The sum from n=1 to infinity}((-1)^n(2)^{n-1})/((2)^{n+1)(n)^{1/3 of} is converges

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sum from n=1 to infinity}((-1)^n(2)^{n-1})/((2)^{n+1)(n)^{1/3 of}

n = 1 \sum \infty \frac{( - 1 ) ^{n} ( 2 ) ^{n - 1}}{( 2 ) ^{n + 1} ( n ) ^{\frac{1}{3}}}

converges

Solution steps

n = 1 \sum \infty \frac{( - 1 ) ^{n} \cdot 2 ^{n - 1}}{2 ^{n + 1} n ^{\frac{1}{3}}}

Apply Alternating Series Test:

converges

= converges

\frac{2 x - 6}{x ^{2} + 9}

y^{^{'}} + \frac{4 y}{x} + \frac{4}{x} = 0, y (1) = - 2

\frac{d}{d x} (3 \cdot e^{\frac{x ^{3}}{9}})

x \to 0 lim (\frac{x}{x})

\int (9 x - 2)^{- 3} d x

What is the sum from n=1 to infinity}((-1)^n(2)^{n-1})/((2)^{n+1)(n)^{1/3 of} ?
The sum from n=1 to infinity}((-1)^n(2)^{n-1})/((2)^{n+1)(n)^{1/3 of} is converges