What is the sum from n=1 to infinity of (1/7)^{n-1} ?

The sum from n=1 to infinity of (1/7)^{n-1} is converges

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sum from n=1 to infinity of (1/7)^{n-1}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{( 0.5 ) ^{n}}{n}

n = 1 \sum \infty 4 (0.9)^{n - 1}

n = 10 \sum \infty \frac{3}{n ^{4.6}}

n = 1 \sum \infty \frac{1}{2 n + 5}

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

What is the sum from n=1 to infinity of (1/7)^{n-1} ?
The sum from n=1 to infinity of (1/7)^{n-1} is converges