What is the sum from n=1 to infinity of 1/(n^{2-1)} ?

The sum from n=1 to infinity of 1/(n^{2-1)} is diverges

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sum from n=1 to infinity of 1/(n^{2-1)}

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

diverges

Solution steps

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

Simplify

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

Apply Cauchy's Convergence Condition:

diverges

= diverges

n = 600 \sum \infty \frac{1}{n}

n = 1 \sum \infty \frac{n}{e ^{7 n}}

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

n = 1 \sum \infty 8 (0.1)^{n - 1}

n = 0 \sum \infty 4 + \frac{5}{n ^{3}}

What is the sum from n=1 to infinity of 1/(n^{2-1)} ?
The sum from n=1 to infinity of 1/(n^{2-1)} is diverges