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 diverges

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

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

diverges

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}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= (- \frac{1}{3}) (n = 0 \sum \infty (- 27)^{n} - (- 27)^{0})

n = 0 \sum \infty (- 27)^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (ln (\frac{x ^{3} + 1}{x ^{2} + 1}))

\int 3 e^{3 t} d t

\frac{x}{x + 1}

\int \frac{20 x ^{3} - 40 x ^{2} + 3}{x ^{2} - 2 x} d x

f (x) = \frac{x}{x - 6}

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 diverges