What is the sum from n=1 to infinity of (4/5)^{n+1} ?

The sum from n=1 to infinity of (4/5)^{n+1} is converges

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sum from n=1 to infinity of (4/5)^{n+1}

n = 1 \sum \infty (\frac{4}{5})^{n + 1}

converges

Solution steps

n = 1 \sum \infty (\frac{4}{5})^{n + 1}

Apply Series Ratio Test:

converges

= converges

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

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

n = 1 \sum \infty \frac{1}{n 4 ^{n}}

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

x = 1 \sum \infty \frac{1}{x ( x + 1 )}

What is the sum from n=1 to infinity of (4/5)^{n+1} ?
The sum from n=1 to infinity of (4/5)^{n+1} is converges