What is the sum from n=1 to infinity of 4(-1/7)^{4n} ?

The sum from n=1 to infinity of 4(-1/7)^{4n} is converges

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sum from n=1 to infinity of 4(-1/7)^{4n}

n = 1 \sum \infty 4 (- \frac{1}{7})^{4 n}

converges

Solution steps

n = 1 \sum \infty 4 (- \frac{1}{7})^{4 n}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty (- \frac{1}{7})^{4 n}

Apply Series Ratio Test:

converges

= 4 converges

= converges

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x \to 2 + lim (\frac{x}{x - 3})

What is the sum from n=1 to infinity of 4(-1/7)^{4n} ?
The sum from n=1 to infinity of 4(-1/7)^{4n} is converges