What is the sum from n=0 to infinity of-6(-1/3)^{2n} ?

The sum from n=0 to infinity of-6(-1/3)^{2n} is converges

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sum from n=0 to infinity of-6(-1/3)^{2n}

n = 0 \sum \infty - 6 (- \frac{1}{3})^{2 n}

converges

Solution steps

n = 0 \sum \infty - 6 (- \frac{1}{3})^{2 n}

Apply the constant multiplication rule:

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

= - 6 \cdot n = 0 \sum \infty (- \frac{1}{3})^{2 n}

Apply Series Ratio Test:

converges

= - 6 converges

= converges

f (x) = 2 x^{3} ln (x)

\frac{d}{d x} ((2 x^{2} - 4)^{3})

\int sin (3 x) sin (n x) d x

(1 - t) (3 + t^{2})^{- 1}

y = \frac{sin ( x )}{x}

What is the sum from n=0 to infinity of-6(-1/3)^{2n} ?
The sum from n=0 to infinity of-6(-1/3)^{2n} is converges