What is the sum from n=3 to infinity of 2/(nln^2(n)) ?

The sum from n=3 to infinity of 2/(nln^2(n)) is converges

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sum from n=3 to infinity of 2/(nln^2(n))

n = 3 \sum \infty \frac{2}{n ln ^{2} ( n )}

converges

Solution steps

n = 3 \sum \infty \frac{2}{n ln ^{2} ( n )}

Apply the constant multiplication rule:

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

= 2 \cdot n = 3 \sum \infty \frac{1}{n ln ^{2} ( n )}

Apply Series Integral Test:

converges

= 2 converges

= converges

n = 0 \sum \infty - \frac{4}{e ^{n}}

n = 0 \sum \infty \frac{1}{e ^{4 n}}

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

k = 0 \sum \infty (- \frac{1}{5})^{k}

n = 1 \sum \infty \frac{8 - 5}{n ^{2}}

What is the sum from n=3 to infinity of 2/(nln^2(n)) ?
The sum from n=3 to infinity of 2/(nln^2(n)) is converges