Question

the recursive sequence is defined as, a1= 9, a2=6, an+1=√an-1+√an , n≥2. Show that the sequence is bounded and strictly decreasing. Find its limit.

Answer 1

Question 1.

The recursive sequence is defined as $a_{1}=9$, $a_{2}=6$, $a_{n+1}=\sqrt{a_{n-1}}+\sqrt{a_{n}}$ , $n>2$. Show that the sequence is bounded and strictly decreasing. Find its limit.

Solution. Note that $a_{3}=\sqrt{a_{2}}+\sqrt{a_{1}}=\sqrt{6}+\sqrt{9}=3+\sqrt{6}<\sqrt{9}+\sqrt{9}=3+3=6=a_{2}$ and $a_{3}=3+\sqrt{6}>3+\sqrt{1}=3+1=4$. Prove by induction that $a_{n}$ is strictly decreasing and bounded below by 4. The base: $a_{1}>a_{2}>a_{3}>1$. Suppose that $1<a_{k}<a_{k-1}$ for all $k\leq n$, where $n\geq 3$. Consider $k=n+1$. Using the recursive formula we see that

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

By inductive hypothesis $1<a_{n}<a_{n-1}$ and $a_{n-2}>a_{n-1}>1$, therefore,

$\sqrt{\frac{a_{n}}{a_{n-1}}}<1,\ \ \sqrt{\frac{a_{n-2}}{a_{n-1}}}>1,$

and hence

$\frac{1+\sqrt{\frac{a_{n}}{a_{n-1}}}}{\sqrt{\frac{a_{n-2}}{a_{n-1}}}+1}<\frac{1+1}{1+1}=1.$

This means that $a_{n+1}<a_{n}$. Furthermore,

$a_{n+1}=\sqrt{a_{n-1}}+\sqrt{a_{n}}>\sqrt{4}+\sqrt{4}=2+2=4.$

Thus, $a_{n}$ is strictly decreasing and bounded below by 4, therefore, we conclude that it has a limit $a\geq 4$. Taking the recursive formula $a_{n+1}=\sqrt{a_{n-1}}+\sqrt{a_{n}}$ and passing to the limit when $n\to\infty$, we get

$a=\sqrt{a}+\sqrt{a}\Leftrightarrow a=2\sqrt{a}\Leftrightarrow\sqrt{a}=2\Leftrightarrow a=4.$

Here we used the fact that $a>0$.

Answer: $\lim_{n\to\infty}a_{n}=4$. $\Box$