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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty n e^{2 n}

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

n = 0 \sum \infty (\frac{n + 5}{n})^{n}

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

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

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