Sequences

At a Glance

Frequency: 2 sub-parts across 2 of 13 years (2016, 2017)
Priority tier: T3
Marks (count): 10 (2)
Average solve time: ~10 min
Difficulty mix: easy 1, medium 1
Section: A | Dominant type: proof

Why This Chapter Matters

Both appearances are in Section A Q1 (compulsory, 10 marks), so these questions are answered under time pressure with no choice. The 2017 question is a clean drill of the Monotone Convergence Theorem on a single recursive sequence — the most tested technique for sequences in UPSC. The 2016 question is harder: two coupled mean-recursion sequences that require induction on a four-term chain and then a limit-equalisation argument. Mastering both templates gives reliable 10-mark scores on any recursive-sequence question.

Minimum Theory

Monotone Convergence Theorem (MCT). A sequence $(x_n)$ of real numbers converges if it is (i) monotone (increasing or decreasing) and (ii) bounded (above if increasing, below if decreasing). An increasing bounded sequence converges to its supremum; a decreasing bounded sequence converges to its infimum.

Standard proof skeleton for a recursive sequence $x_{n+1} = g(x_n)$ . (1) Prove $x_n$ lies in some interval $[a, b]$ by induction (boundedness). (2) Prove $x_{n+1} - x_n$ has constant sign by induction (monotonicity). (3) Apply MCT to get limit $L$ . (4) Pass to the limit in $L = g(L)$ (using continuity of $g$ ) and solve for $L$ . (5) Discard any extraneous root using the bounds from step (1).

Nested sequences and mean inequalities. For positive $x < y$ , $\mathrm{GM}(x,y) = \sqrt{xy} \in (x,y)$ and $\mathrm{HM}(x,y) = 2/(1/x+1/y) \in (x,y)$ . Coupled recursions built on GM and HM produce interlaced, monotone sequences that converge to a common limit, identified by passing to the limit in the product relation.

$Left: a monotone increasing bounded sequence converging to \sup (horizontal dashed line). Right: two interlaced sequences x_n \nearrow L and y_n \searrow L forming nested intervals; the common limit is located by passing to the limit in the product relation x_n^2 = x_{n-1} y_{n-1}.$

Question Archetypes

Archetype	Recognition
monotone-convergence	single recursive sequence $x_{n+1} = g(x_n)$ ; prove convergence
nested-sequences	two coupled sequences defined by mean recursions; prove nesting and common limit

monotone-convergence (1 question; 2017)

Recognition Cues — A single recursion of the form $x_{n+1} = g(x_n)$ with an explicit starting value. The question says “show convergent” without naming the limit — you must prove it exists first. Any explicit root or bound in $g$ (e.g., $\sqrt{\cdot + 20}$ , a fixed point near $5$ ) is a hint for the boundedness argument.

Solution Template

Positivity / well-definedness: show $x_n > 0$ (or stays in the natural domain) by induction.
Find the fixed points $L$ satisfying $L = g(L)$ ; these suggest the natural bound.
Boundedness by induction: show $x_n < L^*$ (the admissible fixed point) for all $n$ .
Monotonicity: compare $x_{n+1}^2$ to $x_n^2$ (or use $x_{n+1} - x_n$ directly), factor, and use the bound from step 3 to determine the sign.
Apply MCT: bounded monotone sequence $\Rightarrow$ convergent; let $L = \lim x_n$ .
Find $L$ : pass to the limit in $L = g(L)$ , solve, discard inadmissible roots by the positivity / sign constraints.

Worked Example

2017 Paper 2, 2017-P2-Q1a (10 marks)

Let $x_1 = 2$ and $x_{n+1} = \sqrt{x_n + 20}$ , $n = 1, 2, 3, \ldots$ . Show that the sequence $x_1, x_2, x_3, \ldots$ is convergent.

Step 1 — Positivity. $x_1 = 2 > 0$ . If $x_n > 0$ then $x_n + 20 > 0$ , so $x_{n+1} = \sqrt{x_n + 20} > 0$ . By induction $x_n > 0$ for all $n$ .

Step 2 — Bounded above by $5$ . Claim $x_n < 5$ for all $n$ . Base: $x_1 = 2 < 5$ . Inductive step: if $x_n < 5$ then

$x_{n+1} = \sqrt{x_n + 20} < \sqrt{5 + 20} = \sqrt{25} = 5.$

Step 3 — Strictly increasing. For $0 < x_n < 5$ , compare squares:

$x_{n+1}^2 - x_n^2 = (x_n + 20) - x_n^2 = -(x_n^2 - x_n - 20) = -(x_n - 5)(x_n + 4).$

Since $x_n < 5$ , we have $x_n - 5 < 0$ and $x_n + 4 > 0$ , so the product is negative and

$x_{n+1}^2 - x_n^2 = -(x_n - 5)(x_n + 4) > 0 \implies x_{n+1} > x_n.$

Step 4 — Apply MCT and find the limit. $(x_n)$ is increasing and bounded above by $5$ , so by the Monotone Convergence Theorem it converges to some $L \le 5$ . Since $x_n \ge x_1 = 2$ , we have $L \ge 2$ . Taking limits in $x_{n+1} = \sqrt{x_n + 20}$ :

$L = \sqrt{L + 20} \implies L^2 = L + 20 \implies (L-5)(L+4) = 0.$

So $L = 5$ or $L = -4$ . Since $L \ge 2 > 0$ , discard $L = -4$ .

$\boxed{(x_n) \text{ converges and } \lim_{n \to \infty} x_n = 5.}$

Common Traps

Establish convergence via MCT before solving $L = \sqrt{L+20}$ . Writing the fixed-point equation first and then claiming “so the limit is $5$ ” earns no marks — the limit exists only because the sequence is bounded and monotone.
The spurious root $L = -4$ is discarded because $L \ge x_1 = 2 > 0$ ; state this explicitly.
The cleanest monotonicity argument uses $x_{n+1}^2 - x_n^2 = -(x_n - 5)(x_n + 4)$ , which reuses the bound $x_n < 5$ established in the previous step.

nested-sequences (1 question; 2016)

Recognition Cues — Two sequences defined by coupled mean-type recursions (e.g., GM and HM of the previous pair). The question asks to prove a four-term chain inequality $x_{n-1} < x_n < y_n < y_{n-1}$ and then deduce a common limit. The key inequalities are GM or HM strictly between the two terms.

Solution Template

Establish positivity of all terms by induction.
State the inductive hypothesis $P(n): x_n < y_n$ . Prove the base case.
In the inductive step, prove the four sub-inequalities (i) $x_{n-1} < x_n$ , (ii) $y_n < y_{n-1}$ , (iii) $x_n < y_n$ ; establish the hinge $x_n < y_{n-1}$ early as it is reused in both (ii) and (iii).
Apply MCT: $(x_n)$ increasing and bounded above (by $y_1$ ); $(y_n)$ decreasing and bounded below (by $x_1$ ). Both converge.
Equate the limits: pass to the limit in the product relation (e.g., $x_n^2 = x_{n-1} y_{n-1}$ ) to get $a^2 = ab$ ; cancel $a > 0$ to obtain $a = b$ .
Locate the common limit using the strict bounds $x_1 < L < y_1$ .

Worked Example

2016 Paper 2, 2016-P2-Q1c (10 marks)

Two sequences $\{x_n\}$ and $\{y_n\}$ are defined inductively: $x_1 = \tfrac{1}{2}$ , $y_1 = 1$ , $x_n = \sqrt{x_{n-1} y_{n-1}}$ , and $\tfrac{1}{y_n} = \tfrac{1}{2}\!\left(\tfrac{1}{x_n} + \tfrac{1}{y_{n-1}}\right)$ for $n \ge 2$ . Prove $x_{n-1} < x_n < y_n < y_{n-1}$ and deduce both sequences converge to the same limit $l$ with $\tfrac{1}{2} < l < 1$ .

Step 1 — Positivity. $x_1, y_1 > 0$ . If $x_{n-1}, y_{n-1} > 0$ then $x_n = \sqrt{x_{n-1} y_{n-1}} > 0$ and $1/y_n$ is a positive average of positives, so $y_n > 0$ . By induction all terms are positive.

Step 2 — Inductive proof of the chain; hypothesis $P(n): x_n < y_n$ .

Base $P(1)$ : $x_1 = \tfrac{1}{2} < 1 = y_1$ . Assume $P(n-1): x_{n-1} < y_{n-1}$ with both positive.

(i) $x_{n-1} < x_n$ . $x_n = \sqrt{x_{n-1} y_{n-1}} > \sqrt{x_{n-1}^2} = x_{n-1}$ since $y_{n-1} > x_{n-1}$ .

(Hinge) $x_n < y_{n-1}$ . $x_n = \sqrt{x_{n-1} y_{n-1}} < \sqrt{y_{n-1}^2} = y_{n-1}$ , i.e., $x_n < y_{n-1}$ .

(ii) $y_n < y_{n-1}$ . From the hinge, $1/x_n > 1/y_{n-1}$ , so

$\frac{1}{y_n} = \frac{1}{2}\!\left(\frac{1}{x_n} + \frac{1}{y_{n-1}}\right) > \frac{1}{2}\cdot \frac{2}{y_{n-1}} = \frac{1}{y_{n-1}} \implies y_n < y_{n-1}.$

(iii) $x_n < y_n$ (proves $P(n)$ ). $y_n$ is the harmonic mean of $x_n$ and $y_{n-1}$ ; since $x_n < y_{n-1}$ (hinge), the HM lies strictly between them:

$\frac{1}{y_n} = \frac{1}{2}\!\left(\frac{1}{x_n} + \frac{1}{y_{n-1}}\right) < \frac{1}{2}\cdot\frac{2}{x_n} = \frac{1}{x_n} \implies y_n > x_n.$

$\boxed{x_{n-1} < x_n < y_n < y_{n-1} \quad (n = 2, 3, 4, \ldots).}$

Step 3 — Convergence.

$(x_n)$ is strictly increasing and bounded above by $y_1 = 1$ (since $x_n < y_n < \cdots < y_1$ ). By MCT, $x_n \to a$ . $(y_n)$ is strictly decreasing and bounded below by $x_1 = \tfrac{1}{2}$ . By MCT, $y_n \to b$ .

Step 4 — Common limit.

Pass to the limit in $x_n^2 = x_{n-1} y_{n-1}$ :

$a^2 = \lim x_n^2 = \lim x_{n-1} y_{n-1} = a \cdot b.$

Since $a \ge x_1 = \tfrac{1}{2} > 0$ , cancel $a$ : $a = b$ . Call the common value $l$ .

Step 5 — Locate $l$ .

$(x_n)$ strictly increases from $x_1 = \tfrac{1}{2}$ , so $l > \tfrac{1}{2}$ . $(y_n)$ strictly decreases from $y_1 = 1$ , so $l < 1$ .

$\boxed{x_n \to l, \quad y_n \to l, \qquad \tfrac{1}{2} < l < 1.}$

Common Traps

The hinge inequality $x_n < y_{n-1}$ (from $x_n = \sqrt{x_{n-1} y_{n-1}} < y_{n-1}$ ) is used in both step (ii) and step (iii); establish it once and refer back.
To show $a = b$ , use the product relation $x_n^2 = x_{n-1} y_{n-1}$ (clean), not the harmonic relation (messy). The cancellation $a^2 = ab \Rightarrow a = b$ requires $a > 0$ , which follows from $a \ge x_1 > 0$ .
The outer bounds must be strict: $l \in (\tfrac{1}{2}, 1)$ not $[\tfrac{1}{2}, 1]$ , justified because the very first step $n = 2$ already moves both sequences inward.

Marks-Aware Writing

10-mark answer (monotone-convergence): Include all four steps — positivity (brief), boundedness (induction), monotonicity (sign of $x_{n+1}^2 - x_n^2$ ), apply MCT then solve $L^2 = L + 20$ , discard $L = -4$ . Each step needs one or two lines of algebra.

10-mark answer (nested-sequences): The induction is the bulk. Set up the hypothesis clearly, prove the base case, then in the inductive step establish the hinge $x_n < y_{n-1}$ before attempting (ii) or (iii). The limit argument (equating $a = b$ via the product relation) is the second major piece and must use MCT by name.

Practice Set

2017-P2-Q1a (10 m) — — Hint: find the fixed point $L = 5$ first as your bound; then show the sequence is increasing via $x_{n+1}^2 - x_n^2 = -(x_n-5)(x_n+4)$ .
2016-P2-Q1c (10 m) — — Hint: the hinge $x_n < y_{n-1}$ is the first thing to establish in the inductive step; use the product relation (not the harmonic one) to equate limits.