What is the sum from n=1 to infinity of 4/(nln(8n)) ?

The sum from n=1 to infinity of 4/(nln(8n)) is diverges

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sum from n=1 to infinity of 4/(nln(8n))

n = 1 \sum \infty \frac{4}{n ln ( 8 n )}

diverges

Solution steps

n = 1 \sum \infty \frac{4}{n ln ( 8 n )}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 4 \cdot n = 1 \sum \infty \frac{1}{n ln ( 8 n )}

Apply Series Integral Test:

diverges

= 4 diverges

= diverges

ln (x^{4})

f (x) = \frac{sec ( x )}{1 + tan ( x )}

\int_{0}^{1} (u + 4) (u - 5) d u

x = 1, x + y = 6

\frac{e ^{x} - ( 1 + x )}{7 x ^{2}}

What is the sum from n=1 to infinity of 4/(nln(8n)) ?
The sum from n=1 to infinity of 4/(nln(8n)) is diverges