What is the sum from n=1 to infinity of ln(n^2+2n) ?

The sum from n=1 to infinity of ln(n^2+2n) is diverges

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sum from n=1 to infinity of ln(n^2+2n)

n = 1 \sum \infty ln (n^{2} + 2 n)

diverges

Solution steps

n = 1 \sum \infty ln (n^{2} + 2 n)

Multiply by the conjugate of

n^{2} + 2 n : \frac{n ^{4} - 4 n ^{2}}{n ^{2} - 2 n}

= n = 1 \sum \infty ln (\frac{n ^{4} - 4 n ^{2}}{n ^{2} - 2 n})

\frac{n ^{4} - 4 n ^{2}}{n ^{2} - 2 n} = n (n + 2)

= n = 1 \sum \infty ln (n (n + 2))

Apply Series Divergence Test:

diverges

= diverges

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What is the sum from n=1 to infinity of ln(n^2+2n) ?
The sum from n=1 to infinity of ln(n^2+2n) is diverges