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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Alternating Series Test:

converges

= 2 converges

= converges

x \to 6 lim (8)

\frac{d}{d θ} (cos (2 θ))

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

\frac{d}{d x} (cosh (e^{x}))

\frac{d y}{d x} + 3 y = 70

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