← 2021 Paper 2

UPSC 2021 Maths Optional Paper 2 Q1b — Step-by-Step Solution

10 marks · Section A

Uniform convergence of series · Real Analysis · asked 3× in 13 yrs · Read the full method →

Question

Test uniform convergence of $x^4+\dfrac{x^4}{1+x^4}+\dfrac{x^4}{(1+x^4)^2}+\cdots$ on $[0,1]$.

Technique

Compute pointwise sum (geometric series); show discontinuity at $x=0$; conclude no uniform convergence.

Solution

Setup. Series $\sum_{n=0}^\infty u_n(x)$ with $u_n(x)=\dfrac{x^4}{(1+x^4)^n}$.

Step 1 — Pointwise sum

For fixed $x\in[0,1]$:

• If $x=0$: all terms are 0; sum = 0.
• If $x\ne 0$: geometric series with first term $x^4$ and ratio $r=1/(1+x^4)\in(0,1)$.

Sum: $S(x)=\dfrac{x^4}{1-r}=\dfrac{x^4}{1-1/(1+x^4)}=\dfrac{x^4(1+x^4)}{x^4}=1+x^4$.

So $S(0)=0$ but $\lim_{x\to 0^+}S(x)=1$. Pointwise limit is discontinuous at $x=0$.

Step 2 — Conclusion on uniform convergence

If the series converged uniformly on $[0,1]$, the limit function would be continuous (since each partial sum is continuous, and uniform limit of continuous functions is continuous).

But $S$ is discontinuous at $0$. So the series does NOT converge uniformly on $[0,1]$.

Answer