What is the sum from n=1 to infinity of 1/(ln(4^n)) ?

The sum from n=1 to infinity of 1/(ln(4^n)) is diverges

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sum from n=1 to infinity of 1/(ln(4^n))

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

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

n = 0 \sum \infty \frac{n ^{2}}{1 + 3 n}

n = 0 \sum \infty (\frac{11}{3})^{n}

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

n = 1 \sum \infty \frac{3}{n ^{2} - 1}

What is the sum from n=1 to infinity of 1/(ln(4^n)) ?
The sum from n=1 to infinity of 1/(ln(4^n)) is diverges