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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

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

x^{2}, 2 - x^{2}

x \to 0 lim (\frac{1}{2})

\int_{0}^{1} x (1 - x)^{8} d x

f (x) = sin (x), 0, \frac{π}{2}

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