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

Question

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?

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

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

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

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

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

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

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

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

Answer 1

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") $\frac{1}{2n}$.

Answer 2

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.

Answer 3

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., $$ \lim\limits_{n\to \infty}\sqrt{n^2+n}-n=\lim\limits_{n\to \infty}\frac{1}{\sqrt{1+\frac{1}{n}}+1}=\frac{1}{1+1}=\frac{1}{2} $$