Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a general strategy for this? For example I'm working on the limit
$$\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n $$

I have a simple argument to show that this limit is less than or equal to 1/2, but I can't get much further because it's difficult for me to manipulate the square root symbol. Here is the argument:
$\sqrt{n^2 + n} - n \lt \sqrt{n^2 +n + \frac{1}{4}} - n = n+\frac{1}{2} - n = 1/2$

share|improve this question
    
You can try a series expansion. –  Amzoti Feb 19 '13 at 0:06
    
Just write $\sqrt{n^2 + n} = n \sqrt{1 + \frac{1}{n}} \rightarrow n$ –  vonbrand Feb 19 '13 at 1:12

3 Answers 3

up vote 8 down vote accepted

$\begin{align} \sqrt{n^2 + n} - n & = \sqrt{n^2 + n} - n \cdot \frac{\sqrt{n^2 + n} + n}{\sqrt{n^2 + n} +n} \\ & = \frac{n^2+n-n^2}{\sqrt{n^2 + n} + n} \\ & = \frac{1}{\frac{\sqrt{n^2 + n} + n}{n}} \\ & = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1} \rightarrow \frac{1}{2}\\ \end{align}$

share|improve this answer
1  
Can't believe I forgot about that trick! –  Mark Feb 19 '13 at 0:14
5  
@Mark: When in doubt, multiply by $1$, or add $0$. :) –  gnometorule Feb 19 '13 at 0:15

$$\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n =\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n\frac{\sqrt{n^2 + n}+n}{\sqrt{n^2 + n}+n}=...=\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^2 + n}+n}=...=\frac{1}{2}$$

share|improve this answer
    
Just noting that this is the same solution as @gnometorule for anyone who is skimming over the solutions. –  Mark Sep 11 at 6:15

By the bound stated in the problem, it suffices to show that the limit is $\ge \frac{1}{2}$. This proof is a lot worse than the others but maybe a good exercise in mean value theorem which is why I'm writing it up. Intuitively we want a way to estimate how far $\sqrt{n^2 + n}$ is from $n$. But the mean value theorem along with concavity gives exactly the approximation we need.

Let $f_n(x) = \sqrt{n^2 + x}$ $$f_n'(x) = \frac{1}{2}(n^2 + x)^{-\frac{1}{2}}$$ Now use the mean value theorem $$\sqrt{n^2 + n} - n = f_n(n) - f_n(0) = f_n'(x^*)n$$ where $x^* \in (0, n)$ (and I have suppressed dependence of $x^*$ on $n$). Then by the fact that $f_n'$ is decreasing, we get $$ f_n'(n) \lt f_n'(x*)$$ Multiplying by $n$ on both sides $$ n f_n'(n) < n f_n'(x^*) = \sqrt{n^2 + n} - n$$ Substituting $$ \frac{1}{2}\frac{n}{(n^2 + n)^{\frac{1}{2}}} \lt \sqrt{n^2 + n} - n$$ $$ \frac{1}{2}\frac{1}{{(1 + 1/n)}^{\frac{1}{2}}} \lt \sqrt{n^2 + n} - n$$

And actually the upper bound of $1/2$ could have been found using the same method.

share|improve this answer
    
I think this is essentially the strategy suggested by @Amzoti with expansion being just the first order taylor approximation = MVT. –  Mark Sep 11 at 6:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.