What is the sum from k=2 to infinity of (3/5)^{3k} ?

The sum from k=2 to infinity of (3/5)^{3k} is converges

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sum from k=2 to infinity of (3/5)^{3k}

k = 2 \sum \infty (\frac{3}{5})^{3 k}

converges

Solution steps

k = 2 \sum \infty (\frac{3}{5})^{3 k}

Apply Series Ratio Test:

converges

= converges

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

k = 1 \sum \infty \frac{1}{2 k}

n = 0 \sum \infty \frac{1}{4} (\frac{3}{4})^{n}

n = 1 \sum \infty 12 sin (\frac{1}{n})

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

What is the sum from k=2 to infinity of (3/5)^{3k} ?
The sum from k=2 to infinity of (3/5)^{3k} is converges