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What are $\limsup{\left(\dfrac{3}{2^{2k+1}}\right)^{1/k}}$ and $\;\;\liminf{\left(\dfrac{3}{2^{2k+1}}\right)^{1/k}}\;?\;\;$ I find both $1/4$. Thank you.

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put on hold as off-topic by 900 sit-ups a day, Claude Leibovici, Miha Habič, 5xum, glace 5 hours ago

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See my answer on your other question and think it through. The hint (and properties of exponents) will get you where you need to go. –  Cameron Buie Dec 27 '12 at 16:41

2 Answers 2

$$\left(\frac{3}{2^{2k+1}}\right)^{1/k}=\left(\frac{3}{2}\right)^{1/k}\frac{1}{2^2}$$

Well, now you only need to know/remember that $\,\sqrt [n] a\xrightarrow [n\to\infty]{}1\,\,\,,\,\,\forall\,a>0$...and you're right: they both are $\,\frac14\,$ (which btw happens iff the sequence converges to the common supremum and infimum limit)

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This may be a useful hint: $${\left(\dfrac{3}{2^{2k+1}}\right)^\frac{1}{k}}=3^\frac{1}{k}(2^{-2k-1})^{\frac{1}{k}}=3^\frac{1}{k}(2^{-2-\frac{1}{k}})=...$$

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