What is the sum from n=1 to infinity of (4/7)^{n-1} ?

The sum from n=1 to infinity of (4/7)^{n-1} is converges

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sum from n=1 to infinity of (4/7)^{n-1}

n = 1 \sum \infty (\frac{4}{7})^{n - 1}

converges

Solution steps

n = 1 \sum \infty (\frac{4}{7})^{n - 1}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty cos (\frac{8}{n})

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

n = 0 \sum \infty (\frac{π}{6})^{n}

n = 0 \sum \infty (- \frac{2}{9})^{n}

n = 3 \sum \infty \frac{5}{n ( n + 1 )}

What is the sum from n=1 to infinity of (4/7)^{n-1} ?
The sum from n=1 to infinity of (4/7)^{n-1} is converges