What is the sum from n=1 to infinity}(-1)^{n-1 of (7/(9^n-1)) ?

The sum from n=1 to infinity}(-1)^{n-1 of (7/(9^n-1)) is converges

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sum from n=1 to infinity}(-1)^{n-1 of (7/(9^n-1))

n = 1 \sum \infty (- 1)^{n - 1} (\frac{7}{9 ^{n} - 1})

converges

Solution steps

n = 1 \sum \infty (- 1)^{n - 1} \frac{7}{9 ^{n} - 1}

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty (- 1)^{n - 1} \frac{1}{9 ^{n} - 1}

Apply Alternating Series Test:

converges

= 7 converges

= converges

\frac{d}{d x} (\frac{x ^{4} + 9}{x ^{2}})

sin (e^{x})

\frac{d}{d x} (3 \frac{8}{x})

\int_{0}^{3} \frac{1}{x ^{2} + 9} d x

x y^{^{'}} + 2 y = 3

What is the sum from n=1 to infinity}(-1)^{n-1 of (7/(9^n-1)) ?
The sum from n=1 to infinity}(-1)^{n-1 of (7/(9^n-1)) is converges