proof of Cauchy’s root test

If for all $n\\geq N$

\\sqrt[n]{a_{n}}<k<1

then

a_{n}<k^{n}<1.

Since $\\sum_{i=N}^{\\infty}k^{i}$ converges so does $\\sum_{i=N}^{\\infty}a_{n}$ by the comparison test. If $\\sqrt[n]{a_{n}}>1$ then by comparison with $\\sum_{i=N}^{\\infty}1$ the series is divergent. Absolute convergence in case of nonpositive $a_{n}$ can be proven in exactly the same way using $\\sqrt[n]{|a_{n}|}$ .

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