What is the sum from n=0 to infinity of 2(1/5)^{n+1} ?

The sum from n=0 to infinity of 2(1/5)^{n+1} is converges

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sum from n=0 to infinity of 2(1/5)^{n+1}

n = 0 \sum \infty 2 (\frac{1}{5})^{n + 1}

converges

Solution steps

n = 0 \sum \infty 2 (\frac{1}{5})^{n + 1}

Apply the constant multiplication rule:

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

= 2 \cdot n = 0 \sum \infty (\frac{1}{5})^{n + 1}

Apply Series Ratio Test:

converges

= 2 converges

= converges

k = 1 \sum \infty \frac{1}{k ( k + 2 )}

n = 1 \sum \infty \frac{n}{3 n ^{3} + 4}

n = 0 \sum \infty (0.03)^{n}

k = 1 \sum \infty (- \frac{6}{7})^{k}

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

What is the sum from n=0 to infinity of 2(1/5)^{n+1} ?
The sum from n=0 to infinity of 2(1/5)^{n+1} is converges