What is the sum from n=1 to infinity of (3/8)^{n-1} ?

The sum from n=1 to infinity of (3/8)^{n-1} is converges

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sum from n=1 to infinity of (3/8)^{n-1}

n = 1 \sum \infty (\frac{3}{8})^{n - 1}

converges

Solution steps

n = 1 \sum \infty (\frac{3}{8})^{n - 1}

Apply Series Ratio Test:

converges

= converges

y = \frac{x ^{2} - 2}{2 x + 1}

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What is the sum from n=1 to infinity of (3/8)^{n-1} ?
The sum from n=1 to infinity of (3/8)^{n-1} is converges