What is the sum from n=0 to infinity of n^{-1} ?

The sum from n=0 to infinity of n^{-1} is diverges

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sum from n=0 to infinity of n^{-1}

n = 0 \sum \infty n^{- 1}

diverges

Solution steps

n = 1 \sum \infty n^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Apply Cauchy's Convergence Condition:

diverges

= diverges

n = 0 \sum \infty 2 - 3 (0.8)^{n}

n = 1 \sum \infty \frac{6}{n ^{2} + n ^{3}}

n = 1 \sum \infty \frac{9}{n ^{2} + 4}

n = 1 \sum \infty \frac{n ^{5}}{8 ^{n}}

n = 0 \sum \infty \frac{n + 1}{2 n + 3}

What is the sum from n=0 to infinity of n^{-1} ?
The sum from n=0 to infinity of n^{-1} is diverges