What is the sum from n=2 to infinity of 6/(nln(n)) ?

The sum from n=2 to infinity of 6/(nln(n)) is diverges

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sum from n=2 to infinity of 6/(nln(n))

n = 2 \sum \infty \frac{6}{n ln ( n )}

diverges

Solution steps

n = 2 \sum \infty \frac{6}{n ln ( n )}

Apply the constant multiplication rule:

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

= 6 \cdot n = 2 \sum \infty \frac{1}{n ln ( n )}

Apply Series Integral Test:

diverges

= 6 diverges

= diverges

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

n = 1 \sum \infty n (- \frac{5}{9})^{n}

n = 0 \sum \infty \frac{1}{n ^{1.5}}

n = 1 \sum \infty \frac{9}{n ^{2} + 49}

n = 0 \sum \infty 3 (\frac{4}{5})^{n}

What is the sum from n=2 to infinity of 6/(nln(n)) ?
The sum from n=2 to infinity of 6/(nln(n)) is diverges