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In many books i see that convergence of a $p$-series when $p>1$ and divergence of a $p$-series when $p$ lies between $0$ and $1$ is proved by either integral test or by Cauchy condensation test. We know Cauchy Criterion for convergence of series is the most basic test for convergence of a series. Can it be done by Cauchy Criterion?

Thank You.

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  • $\begingroup$ Do you mean "Cauchy Criterion" $\endgroup$ – Semsem Feb 26 '14 at 15:19
  • $\begingroup$ yes using cauchy criterion $\endgroup$ – joymath Feb 26 '14 at 15:20
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First prove the following inequality: $$ \frac{1}{k^p}\le\frac{1}{p-1}\Bigl(\frac{1}{(k-1)^{p-1}}-\frac{1}{k^{p-1}}\Bigr). $$ Then $$ \sum_{k=m}^n\frac{1}{k^p}\le \frac{1}{p-1}\Bigl(\frac{1}{(m-1)^{p-1}}-\frac{1}{n^{p-1}}\Bigr). $$

Note: this is the integral test in disguise.

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