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

y = x, y = 3 x

\int \frac{x}{( 5 x ^{2} + 2 ) ^{6}} d x

\int_{0}^{1} \frac{1}{1 + x} d x

y^{^{''}} + 4 y^{^{'}} + 5 y = 5 x - 1

\frac{\partial ^{2}}{\partial x \partial y} (y^{4} cos (2 x))

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