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Show that if $p>1$, $\sum\frac{1}{n^{p}}$ converges and if $p<1$ it diverges for $p\in\mathbb{R}^{+}$.

Is there any way to show another series converges or diverges and then use the Comparison Test to prove this?

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This link [math.stackexchange.com/questions/29450/… could interest you. –  Davide Giraudo Jan 21 '12 at 22:03
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You can show this using the integral test: $\sum \frac{1}{n^p} $ behaves as $\int_1^\infty dx/x^p$ which diverges if $p \leq 1$.

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Why does it diverge if $p\leq1$? –  Emir Jan 21 '12 at 22:48
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$\int_1^N \frac{dx}{x}=\log(N)$. This blows up (slowly) as $N\to\infty$. Then you can do $p<1$ in a similar way, or by comparison with the case $p=1$. –  André Nicolas Jan 22 '12 at 1:04
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There are many ways to skin this particular cat. Among methods not already mentioned, perhaps my favorite is the use of Cauchy's Condensation Test. See $\S 4.3$ of these notes for a statement of CCT and its application to convergence of $p$-series.

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