What is the sum from n=1 to infinity of 2(-1/5)^{2n} ?

The sum from n=1 to infinity of 2(-1/5)^{2n} is converges

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sum from n=1 to infinity of 2(-1/5)^{2n}

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty (- \frac{1}{5})^{2 n}

Apply Series Ratio Test:

converges

= 2 converges

= converges

n = 1 \sum \infty π

n = 1 \sum \infty ln (n - 1)

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

k = 0 \sum \infty (0.1)^{k}

n = 0 \sum \infty \frac{2}{7 - 3 n}

What is the sum from n=1 to infinity of 2(-1/5)^{2n} ?
The sum from n=1 to infinity of 2(-1/5)^{2n} is converges