What is the sum from n=1 to infinity of (-6/25)^n ?

The sum from n=1 to infinity of (-6/25)^n is converges

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sum from n=1 to infinity of (-6/25)^n

n = 1 \sum \infty (- \frac{6}{25})^{n}

converges

Solution steps

n = 1 \sum \infty (- \frac{6}{25})^{n}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{4}{3 n + 2}

n = 1 \sum \infty \frac{1}{12 n + 7}

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

n = 2 \sum \infty (\frac{5}{7})^{2 n}

n = 0 \sum \infty \frac{1}{5 ^{n} + 3}

What is the sum from n=1 to infinity of (-6/25)^n ?
The sum from n=1 to infinity of (-6/25)^n is converges