More on Sequences — Question 7

Question 7 – Solution

Step 1: Rationalize the difference.

Multiplying numerator and denominator by the conjugate avoids the indeterminate form $\infty-\infty$: $$a_n=\frac{(n+1)-n}{\sqrt{n+1}+\sqrt n}=\frac1{\sqrt{n+1}+\sqrt n}.$$

Step 2: Find the limit and monotonicity.

The denominator tends to infinity, so $a_n\to0$. It is also positive. Furthermore, $\sqrt{n+2}+\sqrt{n+1}>\sqrt{n+1}+\sqrt n$; taking reciprocals of positive quantities reverses the inequality and gives $a_{n+1}<a_n$. Thus the sequence decreases to zero.

Step 3: Determine the asymptotic gap size.

Compare $a_n$ with $1/(2\sqrt n)$ by taking their ratio: $$\frac{a_n}{1/(2\sqrt n)}=\frac{2\sqrt n}{\sqrt{n+1}+\sqrt n} =\frac2{\sqrt{1+1/n}+1}\longrightarrow1.$$ Because the ratio tends to $1$, $a_n\sim1/(2\sqrt n)$. Therefore neighboring square roots get closer, and their gap is approximately $1/(2\sqrt n)$ for large $n$.