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UPSC 2020 Maths Optional Paper 2 Q1c — Step-by-Step Solution

10 marks · Section A

Cauchy sequences; completeness of R · Real Analysis · asked 3× in 13 yrs · Read the full method →

Question

Prove that the sequence $(a_n)$ satisfying the condition $|a_{n+1}-a_n|\le\alpha|a_n-a_{n-1}|$, $0<\alpha<1$ for all natural numbers $n\ge2$, is a Cauchy sequence.

Technique

Geometric bound on consecutive differences $\Rightarrow$ telescoping + geometric series tail; standard Cauchy $\varepsilon$–$N$ argument. (This is the “contraction” estimate underlying the Banach fixed-point theorem.)

Solution

Step 1 — Bound consecutive differences geometrically

Write $d_n=|a_{n+1}-a_n|$ for $n\ge1$. The hypothesis says $d_n\le\alpha\,d_{n-1}$ for all $n\ge2$. Iterating,

i.e.

Let $D=|a_2-a_1|$. (If $D=0$ then all $d_n=0$ and the sequence is constant from $a_1$, hence trivially Cauchy. Assume $D>0$.)

Step 2 — Bound $|a_m-a_n|$ for $m>n$ via the triangle inequality

For $m>n\ge1$, telescoping and the triangle inequality give

Since $0<\alpha<1$, the finite geometric sum is bounded by the convergent series:

Therefore

Step 3 — $\alpha^{\,n-1}\to0$, so the tail is arbitrarily small

Since $0<\alpha<1$, $\alpha^{\,n-1}\to 0$ as $n\to\infty$. Hence the right-hand side $\dfrac{D\,\alpha^{\,n-1}}{1-\alpha}\to0$.

Formal $\varepsilon$ argument. Let $\varepsilon>0$. Choose $N$ so large that

This is possible because $\alpha^{\,N-1}\to0$; explicitly any

works (note $\log\alpha<0$). Then for all $m>n\ge N$, since $\alpha^{\,n-1}\le\alpha^{\,N-1}$,

(The case $m=n$ is trivial; $m<n$ follows by symmetry.)

Step 4 — Conclusion

For every $\varepsilon>0$ there exists $N$ with $|a_m-a_n|<\varepsilon$ for all $m,n\ge N$. By definition, $(a_n)$ is a Cauchy sequence.

Answer