Prove $\lim_{n\to\infty}\sqrt{(n+a)(n+b)}-n=\frac{a+b}2$

Question

Prove $$\lim_{n\to\infty}\sqrt{(n+a)(n+b)}-n=\frac{a+b}2.$$

I have found the preceding problem in the Calculus book (Bartle & Sherbert) where I am asked to demonstrate the previous limit assuming $a>0$, $b>0$.

I have been trying for an hour, but I cannot see how I can demonstrate it.

Answer 1

Hint. One may write, as $n \to \infty$, \begin{align} \sqrt{(n+a)(n+b)}-n&=\frac{\left(\sqrt{(n+a)(n+b)}-n\right)\left(\sqrt{(n+a)(n+b)}+n\right)}{\sqrt{(n+a)(n+b)}+n}\\\\ &=\frac{(n+a)(n+b)-n^2}{\sqrt{(n+a)(n+b)}+n} \end{align} Hope you can take it from here.

Answer 2

Remember that $$\sqrt{x}-\sqrt{y}= {x-y\over \sqrt{x}+\sqrt{y}}$$ so

$$\lim_{n \to \infty} \sqrt{(n+a)(n+b)} -n = \lim_{n \to \infty} {(n+a)(n+b) -n^2 \over \sqrt{(n+a)(n+b)} +n}$$

$$= \lim_{n \to \infty} {n\Big((a+b)+{ab\over n}\Big) \over n(\sqrt{(1+{a\over n})(1+{b\over n})} +1)}$$

$$=\frac{a+b}{2}$$

Answer 3

HINT

As an alternative by binomial expansion $$(1+x)^n=1+xn+o(x) \quad x\to 0$$

we have

$$\sqrt{(n+a)(n+b)} -n =n\sqrt{1+\frac a n}\sqrt{1+\frac b n} -n =$$$$= n\left(1+\frac a {2n}+o\left(\frac1n\right)\right)\left(1+\frac b {2n}+o\left(\frac1n\right)\right)-n=\ldots$$

Answer 4

You may consider this sequence limit in the context of the AGM inequality, relating the geometric and the arithmetic mean. The identity $$\sqrt{XY}\;=\;\frac{X+Y}2\:-\:\frac12\left(\sqrt X - \sqrt Y\,\right)^2$$ is a "quantitative AGM version" since the last term involving the square measures the spread between the geometric and the arithmetic mean.

Now let $\,X=n+a, Y=n+b,\,$ and subtract $n$ to obtain $$\sqrt{(n+a)(n+b)}-n\;=\;\frac{a+b}2 \:-\:\frac12\left(\sqrt{n+a}-\sqrt{n+b}\,\right)^2$$ The last term is obviously a null sequence, each element of which is negative (if $a\neq b$).

Thus the limit is approached from below.