Prove the limit: $\lim [\sqrt{4n^2 +n} - 2n] = \frac{1}{4}$

Question

Prove the limit: $\lim [\sqrt{4n^2 +n} - 2n] = \frac{1}{4}$

Discussion: Assume that we can make $\big| [\sqrt{4n^2 +n} - 2n]- \frac{1}{4}\big|$to fall down any given number. Given an arbitrarily small $\varepsilon \gt 0$, we assume $$ \big| [\sqrt{4n^2 +n} - 2n] - \frac{1}{4}\big| \lt \varepsilon $$ $$ \big| [\sqrt{4n^2 +n} - 2n]\big| \lt \varepsilon + 1/4$$ Now, we have two problems here, first that we cannot fully isolate $n$ and second $n$ doesn't occur in denominator as in fractional sequences. Without hurting anyone's feelings we would try to solve the second issue first: $\text{Let's irrationalise the denominator of the expression of the given sequence}$ $\frac{ \left(\sqrt{4n^2 +n} - 2n\right) ~\left( \sqrt{4n^2 +n} + 2n\right)}{ \sqrt{4n^2 +n} +2n}$ $\frac{n }{ \sqrt{4n^2 +n} + 2n} = \frac{1}{ \sqrt{4 +1/n} + 2}$

Let's simply our expression $ \big| [\sqrt{4n^2 +n} - 2n] - \frac{1}{4}\big| = \big| \frac{1}{ \sqrt{4 +1/n} - 2} - 1/4\big| = 1/4 - \frac{1}{ \sqrt{4 +1/n} - 2}~~~\text{ for all} n \in \mathbf{N}$

Now, it's time for solving the first issue, that is to make ##n## floatable, for that we will estimate our original expression by something bigger than that (this is my official argument, that I claim the following expression to be less that epsilon) $$ 1/4 - \frac{1}{ \sqrt{4 +1/n} - 2} \lt 1/4 - \frac{1}{4 +1/n +2} = 1/4 - \frac{1}{6+ 1/n} \lt \varepsilon $$ $$1/4 - \frac{1}{6+ 1/n} \lt \varepsilon\\ 1/4 - \varepsilon \lt \frac{1}{6+1/n} \\ 6/4 - 6 \varepsilon + 1/n ( 1/4 - \varepsilon) \lt 1\\ 1/n ( 1/4 - \varepsilon) \lt 6 \varepsilon - 1/2 \\ n \gt \frac{2 (1/4 - \varepsilon) }{12 \varepsilon -1}$$

Formal Proof:

For any given arbitrarily small $\varepsilon \gt 0$ take $ N = \frac{2 (1/4 - \varepsilon) }{12 \varepsilon -1}$ $$n \gt N \implies n \gt \frac{2 (1/4 - \varepsilon) }{12 \varepsilon -1}\\ 1/n \lt \frac{12 \varepsilon -1}{2 (1/4 - \varepsilon)}\\ 1/n ( 1/4 - \varepsilon) \lt 6\varepsilon -1/2\\ 6/4 - 6\varepsilon + 1/n(1/4 - \varepsilon) \lt 1 \\ 1/4 - \varepsilon \lt \frac{1}{6 + 1/n}\\ 1/4 - \frac{1}{4 + 1/n +2} \lt \varepsilon\\ 1/4 - \frac{ 1}{\sqrt{ 4 +1/n} +2 } \lt 1/4 - \frac{1}{4 + 1/n +2} \lt \varepsilon\\ 1/4 - \frac{ 1}{\sqrt{ 4 +1/n} +2 } \lt \varepsilon \\ \big| \frac{ 1}{\sqrt{ 4 +1/n} +2 } - 1/4 \big| \lt \varepsilon\\ \big| [\sqrt{4n^2 +n} - 2n] - 1/4 \big| \lt \varepsilon$$

Thus, we can the expression $ \big| [\sqrt{4n^2 +n} - 2n] - 1/4 \big|$ to fall below any given number.

So, basically there were three important steps:

1. First get $n$ in denominator.
2. Estimate the original expression by something else, so as to make $n$ to move freely.
3. Judiciously removing the bars of absolute values.

EDIT: Theo Bandit found the mistake and here is the amend:

2nd expression of 5th para of Discussion section:

$$ \frac{1}{4} - \frac{1}{4 +1/n -2} \lt \varepsilon \\ 1/4 - \frac{1}{2+1/n} \lt \varepsilon \\ n \gt \frac{1/4 - \varepsilon}{1/2+ \varepsilon}$$

So, the correct $N$ is $N = \frac{1/4 - \varepsilon}{1/2+ \varepsilon}$. As all our steps were reversible, we can easily establish the formal proof by this new $N$.

Answer 1

You are overcomplicating... You can just consider an upper bound that is easier to handle. For instance, $$ |\sqrt{4n^2+n}-2n -\frac 14|=\left|\dfrac{4n^2+n-(2n+\frac 14)^2}{\sqrt{4n^2+n}+2n+\frac 14}\right|=\dfrac{1/16}{(\cdots)}< \frac{1}{32n} $$

This way, given $\varepsilon >0$, you can choose $N$ such that $\frac{1}{32N} < \varepsilon$.

Answer 2

It would be much faster if you use the Taylor expansion: $$\sqrt{1+x} = 1+\frac{x}{2}+o(x)\; , \qquad x\to 0.$$ In fact: \begin{align*} \sqrt{4n^2+n}-2n &= 2n\bigg(\sqrt{1+\frac{1}{4n}}-1\bigg) =2n\Bigg(\bigg(1+\frac{1}{8n}+o\Big(\frac{1}{n}\Big)\bigg)-1\Bigg) \\ &= \frac{1}{4}+o(1)\;, \qquad n\to\infty\;. \end{align*}

Or, if you don't know Taylor expansions, rely on $(a+b)\cdot(a-b) = a^2 -b^2$ to obtain: \begin{align*} \sqrt{4n^2+n}-2n &= \frac{\big(\sqrt{4n^2+n}-2n) \cdot \big(\sqrt{4n^2+n}+2n)}{\sqrt{4n^2+n}+2n} \\ &= \frac{4n^2+n-4n^2}{\sqrt{4n^2+n}+2n} = \frac{1}{2} \cdot \frac{1}{\sqrt{1+\frac{1}{4n}}+1} \to \frac{1}{4}\;, \qquad n \to \infty \end{align*}

Answer 3

Given the particular closed form of the sequence elements $$\sqrt{4n^2+n}-2n\;=\; \sqrt{2n\left(2n+\frac12\right)}-2n$$ the Harmonic-Geometric-Arithmetic mean inequalities are apt to obtain a fairly quantitative assertion of the convergence $$\frac1{4+\frac1{2n}} \:=\:\frac2{\frac1{2n}+\frac1{2n+\frac12}}-2n \:<\: \sqrt{2n\left(2n+\frac12\right)}-2n \:<\: \frac{2n\;+\;2n+\frac12}2-2n \:=\: \frac14\,,$$ that is we obtain a lower and an upper limit squeezing the sequence terms. As $\,2n\,$ and $\,2n+\frac12\,$ are never equal, the stronger "$<$" holds instead of "$\leqslant$" .

The above inequality chain may be transformed into the equivalent $$0\:<\: \frac14-\big(\sqrt{4n^2+n}-2n\big)\:<\: \frac1{32n+4}\,,$$ from which given some $\,\varepsilon > 0\,$ a corresponding $N(\varepsilon)$ can be readily read off.