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

The sum from n=1 to infinity of ln((n+3)/n) is diverges

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sum from n=1 to infinity of ln((n+3)/n)

n = 1 \sum \infty ln (\frac{n + 3}{n})

diverges

Solution steps

n = 1 \sum \infty ln (\frac{n + 3}{n})

Expand

\frac{n + 3}{n} : 1 + \frac{3}{n}

= n = 1 \sum \infty ln (1 + \frac{3}{n})

Apply Series Integral Test:

diverges

= diverges

n = 1 \sum \infty - \frac{1}{1 3 ^{n}}

n = 0 \sum \infty \frac{1}{n} + \frac{1}{n + 1}

j = 1 \sum \infty 5^{- 3 j}

n = 0 \sum \infty \frac{1}{n ^{8}}

n = 1 \sum \infty sin^{n} (1)

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