What is the sum from n=4 to infinity of 1/n ?

The sum from n=4 to infinity of 1/n is diverges

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sum from n=4 to infinity of 1/n

n = 4 \sum \infty \frac{1}{n}

diverges

Solution steps

n = 4 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= diverges

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

n = 0 \sum \infty \frac{1}{4} (\frac{3}{4})^{n}

n = 1 \sum \infty 12 sin (\frac{1}{n})

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

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

What is the sum from n=4 to infinity of 1/n ?
The sum from n=4 to infinity of 1/n is diverges