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Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$

Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?

$\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon$

$\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| < \epsilon$

$\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} < \epsilon$

$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} > \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$

$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$

$\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$

$\Rightarrow \sqrt{n} > \frac{1-2 \epsilon}{2}$

$\Rightarrow n > \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$

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    $\begingroup$ You need the implications in the other direction, and $$\frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2\epsilon}{2} \not\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2\epsilon}{2}.$$ $\endgroup$ May 6, 2014 at 12:42
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    $\begingroup$ Btw. unless you are explicitly asked to find $N$ for given $\epsilon$ etc., you should more or less stop after reaching transforming $\sqrt {n^2+n}-n$ to $\frac1{1+\sqrt{1+\frac1n}}$. Then note $\frac1n\to 0$, hence $1+\frac1n\to 1$, hence $\sqrt{1+\frac1n}\to 1$, hence $1+\sqrt{1+\frac1n}\to 2$, hence $\frac1{1+\sqrt{1+\frac1n}}\to\frac12$. $\endgroup$ May 6, 2014 at 12:48

10 Answers 10

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From $a^2-b^2=(a+b)(a-b)$ we have

$$\sqrt{n^2+n}-n=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2+n}+n}= \frac{1}{\sqrt{1+\frac{1}{n}}+1}$$

from which the result follows immediately.

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  • $\begingroup$ How could one even possibly see that $a^2+b^2 = (a+b)(a-b)$ works here! $\endgroup$
    – A name
    Jan 30 at 9:48
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Here is another solution. Remark that $$ \sqrt{n^2+n}-n=\sqrt{n^2+n}-\sqrt{n^2}=\frac{(\sqrt{n^2+n}-\sqrt{n^2})(\sqrt{n^2+n}+\sqrt{n^2})}{\sqrt{n^2+n}+\sqrt{n^2}} $$ Then $$ \sqrt{n^2+n}-n=\frac{n}{\sqrt{n^2+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1}. $$ Since $\lim\limits_{n\to \infty} \sqrt{1+\frac{1}{n}}=1$, for $\delta>0$, exists $N$ ($N>\frac{1}{2\delta+\delta^2}$) such that $1<\sqrt{1+\frac{1}{n}}<1+\delta$, for all $n>N$. Then $$ 2<\sqrt{1+\frac{1}{n}}+1<2+\delta $$ So $$ \frac{1}{2}>\frac{1}{\sqrt{1+\frac{1}{n}}+1}>\frac{1}{2+\delta}. $$ Then $$ \frac{\delta}{4}>\frac{1}{2}-\frac{1}{2+\delta}>\frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1}>0 $$ Let $\varepsilon>0$, define $\delta:=4\varepsilon$. So, for $N>\frac{1}{2\delta+\delta^2}=\frac{1}{8\varepsilon+16\varepsilon^2}$, $$ \varepsilon>\frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1}>0, $$ i.e., $$ \lim\limits_{n\to \infty}\sqrt{n^2+n}-n=\lim\limits_{n\to \infty}\frac{1}{\sqrt{1+\frac{1}{n}}+1}=\frac{1}{1+1}=\frac{1}{2} $$

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Since this post became active after a long period of time and there are many linked questions, here is a generalization for such kinds of limits. Consider the set of functions in the following form: $$ f(x) = \sqrt[n]{(x + a_1)(x+a_2)\cdots (x+a_n)} - x $$ where $$ n,k \in \Bbb N \\ a_k \in \Bbb R $$

Consider the following expansion: $$ a^n – b^n = (a – b)\left(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + \cdots + ab^{n – 2} + b^{n – 1}\right) \tag1 $$

Define a function in the form: $$ g(x) = \sqrt[n]{(x + a_1)(x+a_2)\cdots (x+a_n)} $$

Thus $f(x)$ may be rewritten as:

$$ f(x) = g(x) - x $$

Now using $(1)$ we may rewrite $f(x)$ as: $$ f(x) = \frac{(g(x))^n-x^n}{\sqrt[n]{(g(x))^{n-1}} +\sqrt[n]{(g(x))^{n-2}}x + \sqrt[n]{(g(x))^{n-3}}x^2 + \cdots + \sqrt[n]{g(x)}x^{n-2} +x^{n-1}} \tag2 $$

Now taking the limit of $(2)$ one may obtain: $$ \begin{align*} &\lim_{x\to\infty}\frac{x^{n-1}a_1 + x^{n-1}a_2+ \cdots + x^{n-1}a_n }{x^{n-1}\left(\sqrt[n]{1+o\left({1\over x}\right)+\cdots} + \sqrt[n]{1+o\left({1\over x}\right) +\cdots} +\cdots + 1\right)} \\ = &\lim_{x\to\infty}\frac{a_1 + a_2+ \cdots + a_n}{\sqrt[n]{1+o\left({1\over x}\right)+\cdots} + \sqrt[n]{1+o\left({1\over x}\right) +\cdots} +\cdots + 1} \\ = &\frac{a_1 + a_2 + \cdots + a_n}{n} \tag3 \end{align*} $$

Or summarizing: $$ \boxed{\lim_{x\to\infty}f(x) = \frac{a_1 + a_2 + \cdots + a_n}{n}} $$


Lets now use that result in your limit: $$ \sqrt{n^2 + n} - n = \sqrt{n(n+1)} - n = \sqrt{(n+0)(n+1)} - n $$ Which gives $a_1 = 0$ and $a_2 = 1$ and the highest power is $2$ meaning $n = 2$ in $(3)$. Thus we get: $$ \boxed{\lim_{n\to\infty}x_n = \frac{a_1 + a_2}{2} = \frac{0 + 1}{2} = {1\over 2}} $$

As desired.

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You can simply solve the inequality $\frac{1}{\sqrt{1+\frac{1}{n}}+1}>\frac{1-2\epsilon}{2}$and find pick N = $\frac{(1-2\epsilon)}{8\epsilon^2}$

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In general, be wary of the proofs by implications of that sort -- it is easy to miss a step, or to think an implication is an equivalence. Most of the time, what you want can be written as a succession of (in)equalities, but less tricky to handle.

For here, a way to handle this limit, if you have seen what equivalents or $o(\cdot)$'s are: you can start by writing $$ \sqrt{n^2+n}-n = n\sqrt{1+\frac{1}{n}}-n = n\left(\sqrt{1+\frac{1}{n}} - 1\right) $$ and use the fact that $$ (1+x)^\alpha \operatorname*{=}_{x\to 0} 1+\alpha x + o(x) $$ to show that the term in the parentheses is equivalent to ("behaves like") $\frac{1}{2n}$.

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(Yet another solution - sorry!) $$ \left(1 + \frac1{2n}\right)^2 = 1 + \frac1n + \frac1{4n^2} > 1 + \frac1n, $$ therefore $$ \sqrt{1 + \frac1n} < 1 + \frac1{2n}, $$ therefore $$ \sqrt{n^2 + n} - n = n\left(\sqrt{1 + \frac1n} - 1\right) < \frac12, $$ but also $$ \sqrt{n^2 + n} - n = \frac1{\sqrt{1 + \frac1n} + 1} > \frac1{2\left(1 + \frac1{4n}\right)} > \frac{1 - \frac1{4n}}{2} = \frac12 - \frac1{8n}, $$ therefore $$ \frac12 - \epsilon < \sqrt{n^2 + n} - n < \frac12 \quad \left(\epsilon > 0, \ n \geqslant \frac1{8\epsilon}\right). $$

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Here is a high school level solution using L'Hôpital's rule,

$ \lim{\sqrt{n^2+n} -n} = \infty - \infty $

We can write the limit relation in following manner to be able to use L'Hôpital's rule.

$ \lim{~n(\sqrt{1+\frac{1}{n}} -1)} = \infty \times 0 $

rewriting the expression as:

$\lim{\frac{(\sqrt{1+\frac{1}{n}} -1)}{\frac{1}{n}}} $

now can take the derivatives of numerator and denominator as following and solve the problem:

$ \lim{\frac{\frac{-1}{2n^\frac{3}{2}\sqrt{n+1}}}{\frac{-1}{n^2}}} = \lim~\frac{1}{2\sqrt{1+\frac{1}{n}}}$

where $\lim~\frac{1}{2\sqrt{1+\frac{1}{n}}} = \frac{1}{2}$

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Maybe the answer that gives the direct clue about what is going on is going through a proof that

$$\lim_{n \to +\infty} (\sqrt{n^2+n+c_2}-n)-(\sqrt{n^2+n+c_1}-n)=0$$

or equally

$$\lim_{n \to +\infty} \sqrt{n^2+n+c_2}-\sqrt{n^2+n+c_1}=0$$

Now write this to make it obvious as

$$\lim_{n \to +\infty} \frac{1}{\sqrt{n^2+n+c_1}}(\frac{\sqrt{1+\frac{1}{n}+\frac{c_2}{n^2}}}{\sqrt{1+\frac{1}{n}+\frac{c_1}{n^2}}}-1)=0$$

So we can pick whatever constant we want. We pick $\frac{1}{4}$ and have

$$\lim_{n \to +\infty} \sqrt{n^2+n}-n=\lim_{n \to +\infty} (\sqrt{n^2+n+\frac{1}{4}}-n)=$$ $$\lim_{n \to +\infty} (\sqrt{(n+\frac{1}{2})^2}-n)=\lim_{n \to +\infty} n+\frac{1}{2}-n=\frac{1}{2}$$

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Using the fractional binomial theorem we have $$ \Bigl(1+\frac{1}{n}\Bigr)^{1/2}=1+\frac{1}{2n}+o\Bigl(\frac{1}{n}\Bigr) $$ where $o(1/n)$ denotes a quantity that grows asymptotically slower than $1/n$ as $n\to\infty$. Multiplying through by $n$ yields $$ \sqrt{n^2+n}=n+\frac{1}{2}+o(1), $$ which is equivalent to the limit $$ \lim_{n\to\infty}\sqrt{n^2+n}-n=\frac{1}{2}. $$

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Here is yet another proof. Let $n = u-\frac{1}{2}$. Then we have:

$$\lim_{n \to \infty} \sqrt{n(n+1)} - n$$ $$=\lim_{u \to \infty} \sqrt{\left(u-\frac{1}{2}\right) \left(u+\frac{1}{2}\right)} - u + \frac{1}{2}$$ $$=\lim_{u \to \infty} \sqrt{u^2-\frac{1}{4}} - u + \frac{1}{2}$$ $$=\lim_{u \to \infty} u - u + \frac{1}{2} = \frac{1}{2}$$

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