What is the sum from k=2 to infinity of (1/4)^{2k} ?

The sum from k=2 to infinity of (1/4)^{2k} is converges

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sum from k=2 to infinity of (1/4)^{2k}

k = 2 \sum \infty (\frac{1}{4})^{2 k}

converges

Solution steps

k = 2 \sum \infty (\frac{1}{4})^{2 k}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{9 n}{6 n + 1}

0 = 1 \sum \infty \frac{ln ( 0 )}{0}

n = 1 \sum \infty \frac{n ^{4}}{7 ^{n}}

n = 0 \sum \infty (\frac{5}{4})^{n - 1}

n = 3 \sum \infty \frac{1}{5 ^{n} - 3 ^{n}}

What is the sum from k=2 to infinity of (1/4)^{2k} ?
The sum from k=2 to infinity of (1/4)^{2k} is converges