Uniform convergence of series

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2013, 2015, 2021)
• Priority tier: T3
• Marks (count): 10 (1), 13 (1), 15 (1)
• Average solve time: ~7 min
• Difficulty mix: medium 3
• Section: A | Dominant type: proof

Why This Chapter Matters

Uniform convergence is one of the most tested topics in UPSC real analysis: three questions across 2013, 2015, and 2021, at mark weights from 10 to 15. The three questions cover the full range of techniques — uniform Leibniz test for alternating series, pointwise-only convergence for a geometric series that produces a discontinuous limit, and a subtle series-vs-sequence distinction. Knowing these three patterns cold is worth 38 marks in the last 13 years.

Minimum Theory

Uniform convergence of a series. The series $\sum_{n=1}^\infty f_n(x)$ converges uniformly on $D$ if the sequence of partial sums $S_N(x) = \sum_{n=1}^N f_n(x)$ converges uniformly on $D$: for every $\varepsilon > 0$ there exists $N_0$ (independent of $x$) such that $N \ge N_0 \Rightarrow |S(x)-S_N(x)| < \varepsilon$ for all $x \in D$.

Weierstrass M-test. If there exist constants $M_n \ge 0$ with $|f_n(x)| \le M_n$ for all $x \in D$ and all $n$, and if $\sum M_n < \infty$, then $\sum f_n$ converges uniformly and absolutely on $D$.

Uniform Leibniz test. For an alternating series $\sum (-1)^{n-1} a_n(x)$ with $a_n(x) \ge 0$: if (i) $a_{n+1}(x) \le a_n(x)$ for all $x$ and all $n$, and (ii) $\sup_x a_n(x) \to 0$ as $n \to \infty$, then the series converges uniformly on $D$. The key is that the bound on $a_n(x)$ must go to $0$ with an $x$-independent rate.

Continuity of the uniform limit. If each $f_n$ is continuous on $D$ and $\sum f_n$ converges uniformly on $D$, then the sum $S(x)$ is continuous on $D$. Contrapositively: a discontinuous pointwise limit immediately implies non-uniform convergence.

Question Archetypes

series-uniform-convergence (3 question(s); 2013, 2015, 2021)

Recognition Cues — Three sub-patterns have appeared: (a) alternating series $\sum (-1)^{n-1}/(n+x^2)$ — apply uniform Leibniz; (b) geometric series on a closed interval — compute the pointwise sum and check its continuity; (c) series of the form $\sum nx/(1+n^2 x^2)$ — check pointwise convergence first (it may fail entirely for $x \ne 0$), then address the sequence/series distinction.

Solution Template

1. Identify the series type. Is it alternating? Geometric? A general positive series?
2. Pointwise convergence. Find the pointwise limit $S(x)$ (if it exists). For geometric series, sum the geometric progression explicitly.
3. Test for uniform convergence.
   • Alternating: check uniform decay of $a_n(x)$: find an $x$-independent bound $M_n \to 0$ with $a_n(x) \le M_n$.
   • Geometric/continuous limit test: if the pointwise sum is discontinuous, the convergence is not uniform.
   • M-test: if $\sup_x|f_n(x)|$ is summable, conclude uniform absolute convergence.
4. Test for absolute convergence. Consider $\sum |f_n(x)|$ for a fixed $x$ and compare with a known divergent series (typically harmonic).
5. State the conclusion precisely. Distinguish uniform convergence, absolute convergence, and the domain on which each holds.

Worked Example

2013 Paper 2, 2013-P2-Q2c (13 marks)

Show that the series $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n+x^2}$ is uniformly convergent but not absolutely convergent for all real $x$.

Let $a_n(x) = \dfrac{1}{n+x^2}$ and $u_n(x) = (-1)^{n-1} a_n(x)$.

Part 1 — Uniform convergence.

Apply the uniform Leibniz test:

(i) $a_n(x) \ge 0$ since $n \ge 1$ and $x^2 \ge 0$. ✓

(ii) $a_{n+1}(x) = \dfrac{1}{n+1+x^2} < \dfrac{1}{n+x^2} = a_n(x)$ for all $x$. ✓

(iii) Uniform decay: $0 \le a_n(x) = \dfrac{1}{n+x^2} \le \dfrac{1}{n}$ for all $x \in \mathbb{R}$.

The bound $1/n$ is independent of $x$ and $1/n \to 0$. So $a_n(x) \to 0$ uniformly.

By the uniform Leibniz test, $\sum u_n(x)$ converges uniformly on $\mathbb{R}$. The truncation error satisfies $|S(x) - S_N(x)| \le a_{N+1}(x) \le 1/(N+1) \to 0$ uniformly.

Part 2 — Not absolutely convergent.

For fixed $x$, the absolute-value series is

$\sum_{n=1}^\infty \frac{1}{n+x^2}.$

For $n \ge x^2$, we have $n + x^2 \le 2n$, so $\dfrac{1}{n+x^2} \ge \dfrac{1}{2n}$.

The tail $\sum_{n \ge \lceil x^2 \rceil} \dfrac{1}{2n}$ diverges (harmonic). By comparison, $\sum 1/(n+x^2)$ diverges for every fixed $x$.

$\boxed{\text{Series is uniformly convergent on }\mathbb{R},\text{ but not absolutely convergent for any }x.}$

2015 Paper 2, 2015-P2-Q3b (15 marks)

Test the series $\displaystyle\sum_{n=1}^\infty\frac{nx}{1+n^2 x^2}$ for uniform convergence.

Let $u_n(x) = \dfrac{nx}{1+n^2 x^2}$.

Pointwise convergence. At $x = 0$: all terms are $0$; series converges to $0$.

For $x \ne 0$: as $n \to \infty$, $u_n(x) \sim \dfrac{nx}{n^2 x^2} = \dfrac{1}{nx}$. Since $\sum 1/(nx) = \dfrac{1}{x}\sum 1/n$ diverges, the series diverges for every $x \ne 0$.

Sequence interpretation (pedagogical standard answer).

The question is commonly interpreted as asking about the sequence $\{u_n(x)\}$ (each term), since the series itself has no non-trivial domain of convergence.

• Pointwise limit of $u_n(x)$: for $x \ne 0$, $u_n(x) = \dfrac{nx}{1+n^2 x^2} \to 0$ (denominator grows as $n^2$). At $x=0$, $u_n(0)=0$. So $u_n \to 0$ pointwise on $\mathbb{R}$.
• Maximum of $u_n$: by AM-GM, $1 + n^2 x^2 \ge 2n|x|$, so $|u_n(x)| = \dfrac{n|x|}{1+n^2 x^2} \le \dfrac{1}{2}$, with equality at $x = 1/n$. Hence $\sup_x |u_n(x)| = \dfrac{1}{2}$ for every $n$.
• Since $\sup|u_n(x)| = 1/2 \not\to 0$, the convergence $u_n \to 0$ is not uniform on $\mathbb{R}$.
• On $[a, \infty)$ with $a > 0$: for $n$ large enough that $1/n < a$, the max of $u_n$ on $[a,\infty)$ occurs at $x=a$, where $u_n(a) = na/(1+n^2a^2) \to 0$. Convergence is uniform on $[a,\infty)$ for any fixed $a > 0$.

$\boxed{u_n \to 0 \text{ pointwise; not uniform on }\mathbb{R};\text{ uniform on }[a,\infty) \text{ for any } a>0.}$

2021 Paper 2, 2021-P2-Q1b (10 marks)

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

Write $u_n(x) = \dfrac{x^4}{(1+x^4)^n}$, $n \ge 0$.

Pointwise sum. For $x = 0$: all terms $= 0$; sum $= 0$.

For $x \ne 0$: geometric series with first term $x^4$ and ratio $r = 1/(1+x^4) \in (0,1)$:

$S(x) = \frac{x^4}{1 - 1/(1+x^4)} = \frac{x^4(1+x^4)}{x^4} = 1 + x^4.$

Continuity test. The pointwise sum is $S(0) = 0$ but $\lim_{x \to 0^+} S(x) = 1 \ne 0$. So $S$ is discontinuous at $x = 0$.

Conclusion. Each partial sum $S_N(x)$ is continuous on $[0,1]$. If the convergence were uniform, the limit $S$ would be continuous (uniform limit of continuous functions is continuous). But $S$ is discontinuous at $0$. Therefore the series does not converge uniformly on $[0,1]$.

$\boxed{\text{Series does not converge uniformly on }[0,1]\text{ — the pointwise sum is discontinuous at }x=0.}$

Common Traps

• “Uniform decay” must be $x$-independent. In the 2013 problem, $a_n(x) = 1/(n+x^2) \le 1/n$ works because $x^2 \ge 0$ makes the denominator no smaller than $n$. A bound that depends on $x$ does not establish uniform convergence.
• Series diverges vs. sequence converges. The 2015 question is a classic confusion: the sequence $u_n(x) \to 0$ pointwise, but the series $\sum u_n(x)$ diverges for $x \ne 0$ (terms go to zero as $1/(nx)$, and the harmonic series diverges). Address both interpretations if the question is ambiguous.
• Discontinuous limit immediately kills uniform convergence. In the 2021 problem, computing the pointwise sum and noticing the discontinuity at $x = 0$ is the entire argument. No need for epsilon-delta: continuity of the uniform limit is a theorem.
• “Not absolutely convergent” is a pointwise statement. For a fixed $x$, the absolute series diverges — that is what the question is asking. It is not asking whether the absolute series diverges uniformly.

Marks-Aware Writing

A 13–15 mark answer for a uniform convergence proof must include: (1) an explicit statement of the test being applied (Leibniz uniform version, M-test, or continuity criterion), (2) verification of all hypotheses of the test with inequalities written out, (3) for non-uniform convergence via discontinuity — display the pointwise sum, identify the discontinuity, and cite the theorem. For the absolute-convergence part, show the comparison with a harmonic tail explicitly.

A 10-mark answer (2021 style) requires: compute the geometric sum for $x \ne 0$ and separately at $x=0$; identify the jump; cite the continuity theorem; state the conclusion.

Practice Set

• 2019-P2-Q3a (15 m) — — Hint: identify whether the series is alternating or positive-term; check if the Weierstrass M-test applies or if the limit function detects non-uniform convergence.