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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

m = 2 \sum \infty \frac{4}{3 ^{m}}

n = 1 \sum \infty n e^{(- n^{2})}

n = 2 \sum \infty sin (n)

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

n = - 21 \sum \infty - 126

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