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

The sum from n=0 to infinity of ln((n+1)/n) is diverges

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sum from n=0 to infinity of ln((n+1)/n)

n = 0 \sum \infty ln (\frac{n + 1}{n})

diverges

Solution steps

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

Apply Series Limit Comparison Test:

diverges

= diverges

y^{^{''}} + y = sin (2 t)

x \to 5 lim (- 2 x)

f (x) = 3 x + \frac{2}{x}

x \to 0 lim (5 - \frac{x}{2})

x \to 8 - lim (\frac{x}{x - 8})

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