What is the sum from n=1 to infinity of (4/25)^{n-1} ?

The sum from n=1 to infinity of (4/25)^{n-1} is converges

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sum from n=1 to infinity of (4/25)^{n-1}

n = 1 \sum \infty (\frac{4}{25})^{n - 1}

converges

Solution steps

n = 1 \sum \infty (\frac{4}{25})^{n - 1}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{8}{4 n ^{2} - 1}

n = 0 \sum \infty \frac{2 ^{n}}{n ^{5}}

n = 1 \sum \infty cos^{n} (15)

n = 1 \sum \infty cos^{n} (10)

n = 0 \sum \infty (1 - \frac{2}{n})^{3 n}

What is the sum from n=1 to infinity of (4/25)^{n-1} ?
The sum from n=1 to infinity of (4/25)^{n-1} is converges