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UPSC 2017 Maths Optional Paper 2 Q1a — Step-by-Step Solution

10 marks · Section A

Sequences · Real Analysis · asked 2× in 13 yrs · Read the full method →

Question

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.

Technique

Monotone Convergence Theorem — induction for the bound $x_n<5$ and for monotonicity (via comparing $x_{n+1}^2-x_n^2$), then pass to the limit in the recurrence.

Solution

The standard strategy for a recursively defined real sequence is the Monotone Convergence Theorem: a sequence that is monotone and bounded converges. We show $(x_n)$ is increasing and bounded above (by $5$), then identify the limit.

Step 1 — All terms are positive and well defined

$x_1=2>0$. If $x_n>0$ then $x_n+20>0$, so $x_{n+1}=\sqrt{x_n+20}$ is a well-defined positive real. 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

Hence $x_n<5$ for all $n$ by induction.

Step 3 — Monotone increasing

We show $x_{n+1}>x_n$. Since both sides are positive, compare squares:

For $0<x_n<5$ (Step 2) we have $x_n-5<0$ and $x_n+4>0$, so the product $(x_n-5)(x_n+4)<0$, hence

Thus $(x_n)$ is strictly increasing.

Step 4 — Convergence and 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 $x_n\ge x_1=2>0$, we have $L\ge 2>0$. Taking limits in $x_{n+1}=\sqrt{x_n+20}$ (the square root is continuous) gives

So $L=5$ or $L=-4$; as $L>0$ we discard $L=-4$. Therefore

Answer