What is the sum from n=0 to infinity of 3nln(1+n) ?

The sum from n=0 to infinity of 3nln(1+n) is diverges

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sum from n=0 to infinity of 3nln(1+n)

n = 0 \sum \infty 3 n ln (1 + n)

diverges

Solution steps

n = 0 \sum \infty 3 n ln (1 + n)

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty n ln (1 + n)

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

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

n = 1 \sum \infty (\frac{5}{6})^{4 n - 5}

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

k = 2 \sum \infty \frac{1}{2 k ^{2} + k}

n = 1 \sum \infty (cos (6))^{n}

What is the sum from n=0 to infinity of 3nln(1+n) ?
The sum from n=0 to infinity of 3nln(1+n) is diverges