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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 4 converges

= converges

n = 0 \sum \infty n 2^{- n}

n = 1 \sum \infty 9 (0.4)^{n - 1}

n = 1 \sum \infty \frac{2}{n ^{2} + 9}

k = 1 \sum \infty \frac{k ^{2}}{3 ^{k}}

i = 1 \sum \infty \frac{1}{4 i ^{2} - 1}

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