What is the sum from n=1 to infinity of (-6/7)^{n-1} ?

The sum from n=1 to infinity of (-6/7)^{n-1} is converges

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sum from n=1 to infinity of (-6/7)^{n-1}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{e ^{n}}{n + 1}

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

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

n = 0 \sum \infty \frac{3 n}{n + 2}

n = 1 \sum \infty (cos (9))^{n}

What is the sum from n=1 to infinity of (-6/7)^{n-1} ?
The sum from n=1 to infinity of (-6/7)^{n-1} is converges