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UPSC 2026 Maths Optional Paper 2 Q2b — Step-by-Step Solution

15 marks · Section A

Sequences · Real Analysis · asked 3× in 14 yrs · Read the full method →

Question

Prove that the sequence $(u_n)$ defined by the recursion formula $u_{n+1} = \sqrt{7 + u_n}$, $u_1 = \sqrt{7}$, converges to the positive root of $x^2 - x - 7 = 0$.

Technique

The trigger is a recursion $u_{n+1}=f(u_n)$ with $f$ increasing and a fixed point: the route is the monotone convergence theorem — bounded above $+$ increasing $\Rightarrow$ convergent — and the correct bound to induct on is the fixed point itself, i.e. the positive root $L$ of $x^{2}-x-7=0$, because $L=\sqrt{7+L}$ makes the induction step an identity rather than an estimate. Convergence must be established first; only then may one pass to the limit in the recursion. Squaring the recursion to $u_{n+1}^{2}=7+u_n$ before taking limits removes even the need to invoke continuity of $\sqrt{\cdot\,}$.

Solution

Step 0 — The target number, named once and for all.

The roots of $x^{2}-x-7=0$ are $\dfrac{1\pm\sqrt{29}}{2}$. Their product is $-7<0$, so they have opposite signs; there is therefore exactly one positive root, namely

Two properties of $L$ will be used repeatedly:

Numerically $L=\tfrac{1+\sqrt{29}}{2}\approx 3.1926$, while $u_1=\sqrt7\approx2.6458$.

Step 1 — The sequence is well defined and positive.

$u_1=\sqrt7>0$. If $u_n>0$ then $7+u_n>7>0$, so $\sqrt{7+u_n}$ exists as a real number and $u_{n+1}=\sqrt{7+u_n}>\sqrt7>0$. By induction $u_n>0$ for all $n\ge1$, and the recursion never leaves the reals. (Throughout, $\sqrt{\cdot\,}$ denotes the non-negative square root.)

Step 2 — $(u_n)$ is bounded above by $L$.

Claim: $u_n<L$ for all $n\ge1$.

Base $n=1$: since $L>0$,

Induction step: suppose $u_n<L$. The function $t\mapsto\sqrt{7+t}$ is strictly increasing on $[0,\infty)$, so

By induction $u_n<L$ for every $n$. In particular $(u_n)$ is bounded above, and $0<u_n<L$ for all $n$. $\qquad(2.1)$

(This is exactly why $L$, and not some convenient round number, is the bound to choose: $(0.1)$ turns the induction step into an equality, so nothing has to be estimated.)

Step 3 — $(u_n)$ is monotonically increasing.

Base: $u_2=\sqrt{7+u_1}=\sqrt{7+\sqrt7}$ and $u_1=\sqrt7=\sqrt{7+0}$. Since $\sqrt7>0$ we have $7+\sqrt7>7+0$, and $\sqrt{\cdot\,}$ is strictly increasing, so

Induction step: suppose $u_{n+1}>u_n$. Then $7+u_{n+1}>7+u_n$, and applying the strictly increasing map $t\mapsto\sqrt t$,

Hence $u_{n+1}>u_n$ for all $n\ge1$: the sequence is strictly increasing. $\qquad(3.1)$

Cross-check by the quadratic (independent of the induction above). From the recursion,

By $(2.1)$, $M<0<u_n<L$, so $(u_n-L)<0$ and $(u_n-M)>0$, making the right side positive. Thus $u_{n+1}^{2}>u_n^{2}$, and as both terms are positive, $u_{n+1}>u_n$. ✓ (This version also displays the mechanism: the sign of $q(x)=7+x-x^{2}$ on the interval $(M,L)$ is what drives the sequence upward, and it is the vanishing of $q$ at $L$ that stops it.)

Step 4 — Apply the Monotone Convergence Theorem.

Theorem (Monotone Convergence). A monotonically increasing sequence of real numbers which is bounded above converges, and its limit is its supremum.

By $(3.1)$ the sequence is increasing and by $(2.1)$ it is bounded above by $L$. Therefore $(u_n)$ converges. Write

Since $u_n<L$ for all $n$, the supremum satisfies $\ell\le L$; since the sequence increases from $u_1$, we also get

This is the load-bearing step, and it must come before Step 5. The classical error is to begin by writing “let $\ell=\lim u_n$, then $\ell=\sqrt{7+\ell}$”, which presumes precisely what is to be proved. The sequence $u_{n+1}=1+u_n^{2}$, $u_1=1$, satisfies the analogous fixed-point equation $\ell=1+\ell^{2}$ with no real solution at all — because it diverges. Existence of the limit is not a formality; it is the theorem.

Step 5 — Pass to the limit in the recursion (now legitimate).

Square the recursion — this keeps everything polynomial and avoids appealing to the continuity of $\sqrt{\cdot\,}$:

The sequence $(u_{n+1})_{n\ge1}$ is a tail of $(u_n)$ and hence converges to the same limit $\ell$. By the algebra of limits, $u_{n+1}^{2}\to\ell^{2}$ and $7+u_n\to 7+\ell$. Limits of equal sequences are equal, so

Thus $\ell$ is a root of $x^{2}-x-7=0$, so $\ell\in\{L,M\}$.

Step 6 — Identify the root: it is the positive one.

By $(4.1)$, $\ell\ge\sqrt7>0$, whereas $M=\dfrac{1-\sqrt{29}}{2}<0$. Hence $\ell\neq M$, and therefore

the positive root of $x^{2}-x-7=0$. (Two independent reasons exclude $M$: every $u_n>0$ so the limit cannot be negative; and $\ell\le L$ from Step 4 combined with $\ell\in\{L,M\}$ and $\ell>0$ pins $\ell=L$ at once.)

Therefore $(u_n)$ converges, and

Step 7 — Remark: the convergence is geometric (free extra credit).

Using $\sqrt{A}-\sqrt{B}=\dfrac{A-B}{\sqrt A+\sqrt B}$ with $A=7+L$, $B=7+u_n$,

Since $u_{n+1}>u_1=\sqrt7$ (Step 3) and $L=\sqrt{7+L}>\sqrt7$, the denominator exceeds $2\sqrt7=\sqrt{28}>5$, so

the last step because $L-u_1=L-\sqrt7<1$: indeed $q(x)=x^{2}-x-7>0$ for $x>L$, and $q(1+\sqrt7)=(8+2\sqrt7)-(1+\sqrt7)-7=\sqrt7>0$, so $1+\sqrt7>L$. So $u_n\to L$ at least as fast as $5^{-n}$ — which both re-proves convergence (by the sandwich theorem, independently of Step 4) and quantifies it.

Answer