What is the sum from n=1 to infinity of 4(-1/5)^{4n} ?

The sum from n=1 to infinity of 4(-1/5)^{4n} is converges

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sum from n=1 to infinity of 4(-1/5)^{4n}

n = 1 \sum \infty 4 (- \frac{1}{5})^{4 n}

converges

Solution steps

n = 1 \sum \infty 4 (- \frac{1}{5})^{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}{5})^{4 n}

Apply Series Ratio Test:

converges

= 4 converges

= converges

n = 1 \sum \infty 2 (0.9)^{n - 1}

n = 1 \sum \infty (\frac{n}{n + 12})^{n}

n = 1 \sum \infty n^{- 2}

n = 2 \sum \infty \frac{3}{5 ^{n - 1}}

n = 1 \sum \infty \frac{3}{1 + n ^{2}}

What is the sum from n=1 to infinity of 4(-1/5)^{4n} ?
The sum from n=1 to infinity of 4(-1/5)^{4n} is converges