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UPSC 2016 Maths Optional Paper 2 Q1c — Step-by-Step Solution

10 marks · Section A

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

Question

Two sequences $\{x_n\}$ and $\{y_n\}$ are defined inductively by the following:

Prove that $x_{n-1}<x_n<y_n<y_{n-1},\ n=2,3,4,\ldots$ and deduce that both the sequences converge to the same limit $l$, where $\dfrac12<l<1$.

Technique

GM/HM means: GM of $(x,y)$ with $x<y$ lies in $(x,y)$; HM likewise. Induction proves the nested chain $x_{n-1}<x_n<y_n<y_{n-1}$; MCT gives two limits; passing to the limit in $x_n^2=x_{n-1}y_{n-1}$ forces them equal.

Solution

This is a geometric–harmonic mean iteration: $x_n$ is the geometric mean of the previous pair, and $y_n$ is the harmonic mean of $x_n$ and $y_{n-1}$. The two key inequalities are GM $>$ smaller term and HM $<$ larger term, plus the GM–HM inequality. We prove the chain by induction, then nested intervals give a common limit.

Step 1 — Positivity and the base ordering

All terms are positive: $x_1,y_1>0$; 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 $x_n,y_n>0$ for all $n$.

We first establish, for every $n\ge1$, the statement

Base $P(1)$: $x_1=\tfrac12<1=y_1$. ✓

Step 2 — Inductive step: prove the full chain $x_{n-1}<x_n<y_n<y_{n-1}$

Assume $P(n-1):\ x_{n-1}<y_{n-1}$ (with both positive). We derive the four inequalities at index $n$.

(i) $x_{n-1}<x_n$. Since $x_n=\sqrt{x_{n-1}y_{n-1}}$ and $x_{n-1}<y_{n-1}$,

(ii) $y_n<y_{n-1}$. From $1/y_n=\tfrac12(1/x_n+1/y_{n-1})$, $y_n$ is the harmonic mean of $x_n$ and $y_{n-1}$, so it exceeds neither… more precisely we show $1/y_n>1/y_{n-1}$. Since (by (i) and $x_{n-1}<y_{n-1}$) we have $x_n=\sqrt{x_{n-1}y_{n-1}}<\sqrt{y_{n-1}y_{n-1}}=y_{n-1}$, i.e. $x_n<y_{n-1}$, hence $1/x_n>1/y_{n-1}$. Therefore

(iii) $x_n<y_n$ (this is $P(n)$). $y_n$ is the harmonic mean of $x_n$ and $y_{n-1}$, and $x_n<y_{n-1}$ (shown in (ii)). The harmonic mean of two distinct positive numbers lies strictly between them, so $x_n<y_n<y_{n-1}$. Explicitly,

Combining (i)–(iii):

This also re-proves $P(n):x_n<y_n$, completing the induction.

Step 3 — Convergence to a common limit

From Step 2:

• $(x_n)$ is strictly increasing and bounded above (by $y_1=1$, since $x_n<y_n<y_{n-1}<\cdots<y_1=1$). By the Monotone Convergence Theorem $x_n\to a$.
• $(y_n)$ is strictly decreasing and bounded below (by $x_1=\tfrac12$, since $y_n>x_n>x_{n-1}>\cdots>x_1=\tfrac12$). So $y_n\to b$.

The nested intervals $[x_n,y_n]$ are decreasing with $x_n\le a\le b\le y_n$. To show $a=b$, pass to the limit in the GM relation $x_n^2=x_{n-1}y_{n-1}$:

Since $a\ge x_1=\tfrac12>0$ we may cancel $a$: $a=b$. Call the common value $l$.

Step 4 — Locating the limit: $\tfrac12<l<1$

Because $(x_n)$ strictly increases from $x_1=\tfrac12$ and $(y_n)$ strictly decreases from $y_1=1$:

The outer inequalities are strict ($x_1<x_2$ and $y_2<y_1$ from Step 2 at $n=2$), so

Answer