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Let $a \ge-1, a \in \mathbb{R}$. $(a_n)$ is defined as: $$a_0=a, a_n=\sqrt{2^{n+1}a_{n-1}+4^n}-2^n$$ I notice this can be rewritten as $$a_n=2^n \left( \sqrt[2^n]{a+1}-1 \right)$$ so I have dropped the recurrence. Now, this seems to be a classic problem $2^n \to + \infty, \sqrt[2^n]{a+1}-1 \to 0$. Do you have any suggestions as to how I might rewrite $a_n$ to see what the limit is? The answer might depend on $a$ but I don't see it in this form.

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  • $\begingroup$ Write $a+1 = e^{\log (a+1)}$. $\endgroup$ – Daniel Fischer Nov 9 '16 at 10:58
  • $\begingroup$ I did that, but $e^{{ln(a+1) \over 2^n}} \to 1$. I must be missing something else. $\endgroup$ – Zelazny Nov 9 '16 at 12:41
  • $\begingroup$ The Taylor series of the exponential function. $\endgroup$ – Daniel Fischer Nov 9 '16 at 12:43

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