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

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

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

n = 1 \sum \infty ln (2)

diverges

Solution steps

n = 1 \sum \infty ln (2)

Every infinite sum of a non-zero constant diverges

= diverges

n = 1 \sum \infty sin (\frac{πn}{4})

n = 1 \sum \infty \frac{1}{6 ^{n} - 3}

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

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

n = 0 \sum \infty 5 n + 5

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