What is the sum from n=3 to infinity of 7/(3ln(n)) ?

The sum from n=3 to infinity of 7/(3ln(n)) is diverges

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sum from n=3 to infinity of 7/(3ln(n))

n = 3 \sum \infty \frac{7}{3 ln ( n )}

diverges

Solution steps

n = 3 \sum \infty \frac{7}{3 ln ( n )}

Apply the constant multiplication rule:

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

= \frac{7}{3} \cdot n = 3 \sum \infty \frac{1}{ln ( n )}

Apply Series Integral Test:

diverges

= \frac{7}{3} diverges

= diverges

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

n = 0 \sum \infty 3 (\frac{1}{3})^{n}

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

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

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

What is the sum from n=3 to infinity of 7/(3ln(n)) ?
The sum from n=3 to infinity of 7/(3ln(n)) is diverges