What is the sum from n=0 to infinity of (4/5)^{2n} ?

The sum from n=0 to infinity of (4/5)^{2n} is converges

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sum from n=0 to infinity of (4/5)^{2n}

n = 0 \sum \infty (\frac{4}{5})^{2 n}

converges

Solution steps

n = 0 \sum \infty (\frac{4}{5})^{2 n}

Apply Series Ratio Test:

converges

= converges

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

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

n = 0 \sum \infty \frac{n}{1 + n ^{4}}

n = 1 \sum \infty \frac{n !}{8 ^{n}}

n = 3 \sum \infty n^{- 0.9999}

What is the sum from n=0 to infinity of (4/5)^{2n} ?
The sum from n=0 to infinity of (4/5)^{2n} is converges