UPSC 2024 Maths Optional Paper 2 Q2a — Step-by-Step Solution

15 marks · Section A

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

Question

Using Cauchy’s general principle of convergence, examine the convergence of the sequence $\langle f_n\rangle$ , where $f_n=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots+\dfrac{1}{n!}$ .

Technique

Bound $|f_m-f_n|$ using $k!\ge 2^{k-1}$ to reduce to a geometric tail; then apply the Cauchy criterion.

Solution

Cauchy’s general principle. A sequence $\langle f_n\rangle$ converges iff for every $\varepsilon>0$ there exists $N$ such that $|f_m-f_n|<\varepsilon$ for all $m,n\ge N$ .

Step 1 — Bound $|f_m-f_n|$ for $m>n$ .

|f_m-f_n|=\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\cdots+\frac{1}{m!}.

Since $k!\ge 2^{k-1}$ for all $k\ge 1$ (by induction), we have $1/k!\le 1/2^{k-1}$ :

|f_m-f_n|\le\sum_{k=n+1}^{m}\frac{1}{2^{k-1}}<\sum_{k=n+1}^{\infty}\frac{1}{2^{k-1}}=\frac{1}{2^{n-1}}.

Step 2 — Apply the criterion.

Given $\varepsilon>0$ , choose $N$ such that $1/2^{N-1}<\varepsilon$ . Then for all $m,n\ge N$ (with $m>n$ ):

|f_m-f_n|\le\frac{1}{2^{n-1}}\le\frac{1}{2^{N-1}}<\varepsilon.

By Cauchy’s principle, $\langle f_n\rangle$ is convergent.

Answer

\boxed{\;\langle f_n\rangle\text{ converges.}\;\;(\text{Its limit is }e=\sum_{k=0}^\infty 1/k!.)\;}