I have to find the limit of sequences, and there are two where I would like to use the squeeze theorem but unable to do the algebraic manipulation before I can implement it:

$$a_n= (5+n 4^n)^{1/(2n)}$$

$$b_n= (4n^2+7)^{0.5} -2n$$

Any ideas as to how I could rearrange/get rid of the power for $a_n$?

  • $\begingroup$ Hint: $a_n:=\exp(\log a_n)= \exp((2n)^{-1}\log (5+n4^n))$... $\endgroup$ – Milly Nov 2 '14 at 19:02
  • $\begingroup$ We have not yet looked at anything involving logarithms with limits, and are meant to use rules/theorems such as squeeze theorem, ratio test, sum/product rule to solve the limits $\endgroup$ – MathsUndergrad Nov 2 '14 at 19:04
  • $\begingroup$ Ok, then for $a_n$ use $(4^n)^{1/2n}\leq a_n\leq (2n)^{1/2n} (4^n)^{1/2n}$ $\endgroup$ – Milly Nov 2 '14 at 19:08
  • $\begingroup$ How are you able to prove that holds, I am unsure how you would arrive at those 2 expressions $\endgroup$ – MathsUndergrad Nov 2 '14 at 19:11

rewrite your first term as $2^\cdot n^{\frac{1}{2n}}\left(\frac{5}{n}+1\right)^{1/(2n)}$ and rewrite your second term in the form $\left(\sqrt{4n^2+7}-2n\right)\frac{\sqrt{4n^2+7}+2n}{\sqrt{4n^2+7}+2n}$


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