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

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

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

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

converges

Solution steps

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

Simplify

(- 3)^{- 3 n - 1} : (- 3)^{- 3 n} (- 3)^{- 1}

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

Apply the constant multiplication rule:

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

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

(- 3)^{- 1} = - \frac{1}{3}

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

Simplify

(- 3)^{- 3 n} : (- 27)^{- n}

= (- \frac{1}{3}) \cdot n = 1 \sum \infty (- 27)^{- n}

Apply Series Ratio Test:

converges

= (- \frac{1}{3}) converges

= converges

f (x) = e^{3 x}, (0, 1)

\frac{d}{d x} ((2 x + 5) (x + 1))

\frac{d}{d x} (\frac{4 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}})

y + y^{^{'}} = sin (t)

t^{9} cos (t)

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