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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

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

n = 1 \sum \infty (\frac{2}{n})^{4}

n = 1 \sum \infty \frac{3}{n ^{2} + 3}

n = 0 \sum \infty n^{^{'}} + n

n = 0 \sum \infty n^{^{'}} - n

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