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Please kindly help with this problem on converging sequence.

Question Determine whether the sequence $\bigl(\sqrt{4n^2+n}-2n\bigr)_{n\in\mathbb N}$ converges and if it converges, guess its limits and prove your guess.

Now what I don't really understand is to proof the convergence of the sequence, because it has no limit point, so how do we prove it is convergent, and show me your guess.

Martin Sleziak
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Abubakar M. Shafii
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2 Answers2

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Your best option is to guess the limit. Note that\begin{align*}\sqrt{4n^2+n}-2n&=\frac{\left(\sqrt{4n^2+n}-2n\right)\left(\sqrt{4n^2+n}+2n\right)}{\sqrt{4n^2+n}+2n}\\&=\frac n{\sqrt{4n^2+n}+2n}\\&=\frac1{\sqrt{4+\frac1n}+2}.\end{align*}I suppose that now it won't be difficult for you to guess the limit and to prove that your guess is right.

José Carlos Santos
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  • OK I think I understand now... But if I guess my limit to be $\frac{1}{4}$ , then can you show me how to proof my guess, I tried it using epsilon definition of convergency and I got stucked.. – Abubakar M. Shafii Aug 18 '17 at 15:58
  • @AbubakarShafii $\lim_{n\in\mathbb N}\frac1{\sqrt{4+\frac1n}+2}=\frac1{\lim_{n\in\mathbb N}\sqrt{4+\frac1n}+2}=\frac1{2+2}=\frac14$. – José Carlos Santos Aug 18 '17 at 16:00
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hint just use

$$\sqrt {A}-B=\frac {(\sqrt {A}-B)(\sqrt {A}+B)}{\sqrt {A}+B} $$ $$\frac {A-B^2}{\sqrt {A}+B} $$

and recompute le limit after a simplification.

hamam_Abdallah
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  • OK I think I understand now... But if I guess my limit to be $\frac{1}{4}$, then can you show me how to proof my guess, I tried it using epsilon definition of convergency and I got stucked.. – Abubakar M. Shafii Aug 18 '17 at 12:19