What is the sum from n=0 to infinity of ln(1+n^2) ?

The sum from n=0 to infinity of ln(1+n^2) is diverges

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sum from n=0 to infinity of ln(1+n^2)

n = 0 \sum \infty ln (1 + n^{2})

diverges

Solution steps

n = 0 \sum \infty ln (1 + n^{2})

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{1}{3 ^{n} + n}

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

k = 1 \sum \infty k^{2}

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

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

What is the sum from n=0 to infinity of ln(1+n^2) ?
The sum from n=0 to infinity of ln(1+n^2) is diverges