What is the sum from n=1 to infinity of 3(-1/5)^{3n} ?

The sum from n=1 to infinity of 3(-1/5)^{3n} is converges

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sum from n=1 to infinity of 3(-1/5)^{3n}

n = 1 \sum \infty 3 (- \frac{1}{5})^{3 n}

converges

Solution steps

n = 1 \sum \infty 3 (- \frac{1}{5})^{3 n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty (- \frac{1}{5})^{3 n}

Apply Series Ratio Test:

converges

= 3 converges

= converges

n = 2 \sum \infty \frac{e ^{\frac{1}{n}}}{n}

n = 0 \sum \infty (\frac{4}{3})^{1 - n}

n = 2 \sum \infty \frac{1 ^{n}}{ln ( n )}

n = 1 \sum \infty 10 (- \frac{7}{8})^{n}

n = 0 \sum \infty 4 n

What is the sum from n=1 to infinity of 3(-1/5)^{3n} ?
The sum from n=1 to infinity of 3(-1/5)^{3n} is converges