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For what $p$ values does the series $\sum_{n=1}^\infty \sqrt{n^{p-4}}$ converge?

I think it's $\forall p\in[1,2)$, but I can't show it.

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    $\begingroup$ Recall that $\sum_1^\infty \frac{1}{n^q}$ converges when $q\gt 1$ and diverges otherwise. $\endgroup$ – André Nicolas Aug 16 '15 at 7:09
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Let $p \in \mathbb{R}$. Then $\sum_{n} n^{\frac{p-4}{2}}$ converges if and only if $\frac{p-4}{2} < -1$, that is, $p < 2$.

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$$Hint*\quad :\quad (\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ p } } } ,\quad p\quad is\quad positive\quad ,\quad \quad 0<\quad p\quad \le 1\quad )\quad \rightarrow \quad series\quad is\quad divergent\\ you\quad can\quad reach\quad for\quad the\quad same\quad result\quad by\quad Integeral\quad test\quad $$

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