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Prove using the definition of a limit that $$\displaystyle{\lim_{n\to\infty}\underbrace{\sqrt{n^2+n}-n}_{s_n}}=\frac{1}{2}.$$

Proof:

Let $\epsilon>0$ and $n\in\mathbb{N}$, then $$\left|\sqrt{n^2+n}-n-\frac12\right|\stackrel{(*)}{=}\left|\frac{n-\sqrt{n^2+n}}{2(n+\sqrt{n^2+n})}\right|=\left|\frac{n}{2(n+\sqrt{n^2+n})^2}\right|=\left|\frac{n}{4n^2+4n\sqrt{n^2+n}+2n}\right|<\left|\frac{n}{4n\sqrt{n^2+n}+2n}\right|=\left|\frac{1}{4\sqrt{n^2+n}+2}\right|$$ Since $\forall n\in\mathbb{N}\,\,\sqrt{n^2+n}>\sqrt{n}\,$ then $$\left|\frac{1}{4\sqrt{n^2+n}+2}\right|<\left|\frac{1}{4\sqrt{n}}\right|<\epsilon\quad\text{ if }\quad n>\frac{1}{16\epsilon^2}.$$

Would this be correct?

(*) maybe a simpler way would be to recognize that the numerator is $|s_n|<1?$

DMH16
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  • Abit lost at your first expression after =, why don’t you just multiply the original by an expression to complete the difference of squares (and divide, respectively, to compensate), and then just divide numerator and denominator by $n$? – Makina Dec 15 '18 at 05:49
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    A more direct way is to note that $$\sqrt{n^2+n}-n=\frac1{\sqrt{1+\frac1n}+1}$$ whose limit is direct, as has been explained tons of times on the site (did you search for duplicates before posting this?). – Did Dec 15 '18 at 05:50
  • @Did not sure where that leads me to... I am supposed to use the definition of a limit to prove the statement, not just compute it. I am able to compute the limit, but that's not the scope here – DMH16 Dec 15 '18 at 05:53
  • @Makina that's what I did, but then dividing by $n$ as mentioned by Did does not really help a whole lot, since I am not trying to compute the limit with limit properties, but show that the equality holds using the definition – DMH16 Dec 15 '18 at 05:54
  • Possible duplicate of Prove that lim $(\sqrt{n^2+n}-n) = \frac{1}{2}$ – Nosrati Dec 15 '18 at 09:43
  • @Nosrati No, it’s not. None of the answers (except the second, which seems overly complicated) use the definition of a limit to answer the question. Please read the problem statement above. Also, I am asking if my proof is correct, thus a proof-verification question. – DMH16 Dec 15 '18 at 09:47
  • What have you done in (*) ? – Nosrati Dec 15 '18 at 10:02
  • "not sure where that leads me to" You would see "where this leads you to" if you tried to compute $\sqrt{n^2+n}-n-\frac12$ using the formula in my comment. – Did Dec 15 '18 at 10:07
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    @DMH16 Your algebra and logic are correct, what part of the proof are you uncertain about? Can you justify the use of inequalities to yourself? – DanielV Dec 15 '18 at 11:13

1 Answers1

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We know that $$\sqrt{n^2+n}-n{=(\sqrt{n^2+n}-n){\sqrt{n^2+n}+n\over \sqrt{n^2+n}+n}\\={n\over \sqrt{n^2+n}+n}\\={1\over 1+\sqrt{1+{1\over n}}}}$$therefore$$\left|{1\over 1+\sqrt{1+{1\over n}}}-{1\over 2}\right|{={1\over 2}-{1\over 1+\sqrt{1+{1\over n}}}<\epsilon}$$which means that $${ 1+\sqrt{1+{1\over n}}}<{1\over {1\over 2}-\epsilon}\\\sqrt{1+{1\over n}}<{{1\over 2}+\epsilon\over {1\over 2}-\epsilon}\\{1+{1\over n}}<{{1\over 4}+\epsilon^2+\epsilon\over {1\over 4}+\epsilon^2-\epsilon}\\{1\over n}<{2\epsilon\over {1\over 4}+\epsilon^2-\epsilon}$$therefore by choosing $$n>{{1\over 4}+\epsilon^2-\epsilon\over 2\epsilon}$$or even $n>{1\over 8\epsilon}$ for small enough $\epsilon$ we obtain $$\left|{1\over 1+\sqrt{1+{1\over n}}}-{1\over 2}\right|<\epsilon$$

Mostafa Ayaz
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  • Thanks for the answer. I agree with your work, but would my procedure be correct? My question is whether my proof is correct or not. – DMH16 Dec 15 '18 at 09:51
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    Yes. Quite right to me since despite of correction of your proof (though it is correct), the bound you attained to, overtakes mine :) – Mostafa Ayaz Dec 15 '18 at 11:41