What is the sum from n=100 to infinity of 1/n ?

The sum from n=100 to infinity of 1/n is diverges

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sum from n=100 to infinity of 1/n

n = 100 \sum \infty \frac{1}{n}

diverges

Solution steps

n = 100 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= diverges

n = 2 \sum \infty (- 0.4)^{n}

n = 0 \sum \infty π^{- 3 n}

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

n = 0 \sum \infty 8 (\frac{2}{3})^{n}

n = 0 \sum \infty \frac{2}{n ^{3}}

What is the sum from n=100 to infinity of 1/n ?
The sum from n=100 to infinity of 1/n is diverges