What is the sum from n=1 to infinity of 1/(ln(3^n)) ?

The sum from n=1 to infinity of 1/(ln(3^n)) is diverges

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sum from n=1 to infinity of 1/(ln(3^n))

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

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

k = 2 \sum \infty k e^{- k^{2}}

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

n = 1 \sum \infty \frac{1}{n ln ^{2} ( n )}

n = 0 \sum \infty n^{- s i n (1)}

What is the sum from n=1 to infinity of 1/(ln(3^n)) ?
The sum from n=1 to infinity of 1/(ln(3^n)) is diverges