How to evaluate the limit as $x\to0$ of
$$\biggl(\,\sum_{r=1}^{r=n}r^{1/\sin^2x}\biggr)^{\!\sin^{2}x}$$
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1Any thoughts on how to approach this problem? Evidently $n$ is to be some fixed positive integer. Start by considering $n=1$ if you need a concrete case to suggest approaches. – hardmath Jun 26 '15 at 13:32
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Let $p=\frac{1}{\sin^2 x}$. As $x\to 0^+$, $p\to +\infty$, and: $$ \left(\sum_{r=1}^{n} r^p\right)^{1/p}=\|(1,2,\ldots,n)\|_p \to \max(1,2,\ldots,n)=n.$$
Jack D'Aurizio
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Is it trivial to show that the limit of the $L_p$ norms as $p\to\infty$ is the $L_\infty$ norm? – Braindead Jun 26 '15 at 12:17
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1Found the proof :P http://math.stackexchange.com/questions/326172/the-l-infty-norm-is-equal-to-the-limit-of-the-lp-norms – Braindead Jun 26 '15 at 12:35