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

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

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

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

diverges

Solution steps

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

Apply Series Limit Comparison Test:

diverges

= diverges

n = 1 \sum \infty \frac{5}{n ln ( 3 n )}

n = 1 \sum \infty \frac{9}{n !}

n = 0 \sum \infty \frac{n}{5 ^{n}}

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

n = 0 \sum \infty \frac{n}{1 + 2 n ^{2}}

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