What is the sum from n=1 to infinity of (0.7)^{n-1} ?

The sum from n=1 to infinity of (0.7)^{n-1} is converges

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sum from n=1 to infinity of (0.7)^{n-1}

n = 1 \sum \infty (0.7)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 0. 7^{n - 1}

Apply Series Ratio Test:

converges

= converges

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

n = 0 \sum \infty \frac{e ^{n}}{n ^{4}}

n = 1 \sum \infty \frac{1}{5 n + 3}

n = - \infty \sum \infty n

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

What is the sum from n=1 to infinity of (0.7)^{n-1} ?
The sum from n=1 to infinity of (0.7)^{n-1} is converges