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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

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

n = 0 \sum \infty \frac{1}{n ^{2} - 4}

k = 1 \sum \infty \frac{1}{7 ^{k} + 7}

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

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

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