What is the sum from n=1 to infinity of (0.75)^{n-1} ?

The sum from n=1 to infinity of (0.75)^{n-1} is converges

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sum from n=1 to infinity of (0.75)^{n-1}

n = 1 \sum \infty (0.75)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 0.7 5^{n - 1}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{3 ^{n}}{7 ^{n}}

n = 3 \sum \infty \frac{6}{n ^{3}}

n = 1 \sum \infty 4 (1.3)^{n}

n = 0 \sum \infty n^{^{''}} - n^{^{'}}

n = 0 \sum \infty (\frac{99}{100})^{n}

What is the sum from n=1 to infinity of (0.75)^{n-1} ?
The sum from n=1 to infinity of (0.75)^{n-1} is converges