What is the sum from n=1 to infinity of ln(n/(2n+5)) ?

The sum from n=1 to infinity of ln(n/(2n+5)) is diverges

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sum from n=1 to infinity of ln(n/(2n+5))

n = 1 \sum \infty ln (\frac{n}{2 n + 5})

diverges

Solution steps

n = 1 \sum \infty ln (\frac{n}{2 n + 5})

Apply Series Divergence Test:

diverges

= diverges

\int \frac{x ^{2} - x + 6}{x ^{3} + 3 x} d x

\frac{\partial}{\partial x} (\frac{x \cdot y}{x + y})

\frac{d}{d x} ((\frac{x ^{2} + 1}{x ^{2} - 1})^{8})

\frac{\partial}{\partial x} (x^{5} + x y^{7} + 4)

x \to - 3 - lim (4 x^{2} - 6 x + 3)

What is the sum from n=1 to infinity of ln(n/(2n+5)) ?
The sum from n=1 to infinity of ln(n/(2n+5)) is diverges