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

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

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

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

diverges

Solution steps

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

Apply Series Limit Comparison Test:

diverges

= diverges

n = 2 \sum \infty \frac{6}{n ln ( n )}

n = 1 \sum \infty 4^{\frac{1}{n}}

n = 1 \sum \infty n (- \frac{5}{9})^{n}

n = 0 \sum \infty \frac{1}{n ^{1.5}}

n = 1 \sum \infty \frac{9}{n ^{2} + 49}

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