What is the sum from n=1 to infinity of (4/9)^{n-1} ?

The sum from n=1 to infinity of (4/9)^{n-1} is converges

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sum from n=1 to infinity of (4/9)^{n-1}

n = 1 \sum \infty (\frac{4}{9})^{n - 1}

converges

Solution steps

n = 1 \sum \infty (\frac{4}{9})^{n - 1}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty (3.5)^{- n}

n = 1 \sum \infty \frac{10 n}{4 n + 1}

k = 4 \sum \infty \frac{1}{( k - 3 ) ^{4}}

k = 1 \sum \infty 30 (0.1)^{k}

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

What is the sum from n=1 to infinity of (4/9)^{n-1} ?
The sum from n=1 to infinity of (4/9)^{n-1} is converges