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

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

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

n = 2 \sum \infty 2 n ln (n)

diverges

Solution steps

n = 2 \sum \infty 2 n ln (n)

Apply the constant multiplication rule:

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

= 2 \cdot n = 2 \sum \infty n ln (n)

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

\frac{d}{d x} (\frac{x - 7}{x + 7})

\int 9 x^{\frac{2}{3}} d x

x \to - \infty lim (3 - 2^{x})

ln (\frac{x}{x ^{2} + c})

\int_{0}^{8} \frac{2}{x ^{\frac{2}{3}}} d x

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