← 2015 Paper 2

UPSC 2015 Maths Optional Paper 2 Q3b — Step-by-Step Solution

15 marks · Section A

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

Question

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

Technique

Pointwise limit of general term; max of $u_n(x)$ at $x=1/n$ ($u_n(1/n)=1/2$); Weierstrass M-test fails because $\sup|u_n|=1/2$ doesn’t sum.

Solution

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

Step 1 — Pointwise convergence

Fix $x$. For $x=0$: $u_n(0)=0$ for all $n$; series sum is $0$.

For $x\ne 0$: as $n\to\infty$, $u_n(x)=\dfrac{nx}{1+n^2 x^2}\sim\dfrac{nx}{n^2 x^2}=\dfrac{1}{nx}$.

So $\sum u_n(x)\sim\dfrac{1}{x}\sum\dfrac{1}{n}$, which diverges (harmonic).

So the series diverges for every $x\ne 0$ and converges trivially at $x=0$.

Pointwise: series converges only at $x=0$. Uniform convergence is therefore vacuous on any interval not entirely $\{0\}$.

Wait — let me reconsider. The problem asks for uniform convergence. If pointwise convergence fails on most points, the question becomes more nuanced. Let me re-examine.

Step 2 — Re-examine $u_n$

$u_n(x)=\dfrac{nx}{1+n^2 x^2}$. As a function of $x$, $u_n$ has max at $x=1/n$ where $u_n(1/n)=\dfrac{1}{2}$.

For fixed $x\ne 0$: $|u_n(x)|\sim 1/(n|x|)$ as $n\to\infty$, which is summable… wait, $\sum 1/n$ is not summable; $\sum 1/(n|x|)$ diverges.

So $\sum u_n(x)$ diverges for $x\ne 0$. Series does not converge pointwise on any set containing a nonzero point.

Hmm — let me sanity check with $x=1$: $u_n(1)=n/(1+n^2)$, and $\sum n/(1+n^2)\sim\sum 1/n$ diverges. ✓

So strictly speaking, the series has no domain of pointwise convergence besides $\{0\}$. Uniform convergence on a set requires pointwise convergence first; since pointwise fails for $x\ne 0$, the series cannot converge uniformly on any set containing $x\ne 0$.

Step 3 — Reinterpret the question

The standard UPSC reading of this problem: test for uniform convergence on a specific interval, e.g. $[a,\infty)$ for $a>0$, or examine the sequence $\{u_n(x)\}$ (not the series). Let me check if the original problem might have meant sequence not series — UPSC PDFs sometimes have ambiguous formatting.

If it’s the sequence $u_n(x)$:

• Pointwise limit: for fixed $x\ne 0$, $u_n(x)=nx/(1+n^2 x^2)\to 0$ as $n\to\infty$ (denom grows faster). At $x=0$, $u_n(0)=0$. So $u_n\to 0$ pointwise on $\mathbb R$.
• Uniform convergence on $\mathbb R$: $\sup_x|u_n(x)|=u_n(1/n)=1/2$, which does not go to $0$. So $u_n$ does not converge uniformly on $\mathbb R$.
• On $[a,\infty)$ for $a>0$: $u_n(x)$ has max in this range. For $n$ large, $1/n<a$, so max on $[a,\infty)$ occurs at $x=a$: $u_n(a)=na/(1+n^2 a^2)\to 0$. Convergence on $[a,\infty)$ is uniform.

Given the original problem says “series of functions”, I’ll stick with that interpretation and conclude:

Step 4 — Conclusion (series interpretation)

The series $\sum_{n=1}^\infty u_n(x)$ diverges at every $x\ne 0$ (by limit comparison with $\sum 1/n$). Hence uniform convergence is not meaningful on any non-trivial set.

However, the question is likely intended in the sense of the sequence of partial sums. Alternative reading (sequence): $\{u_n(x)\}$ converges to $0$ pointwise on $\mathbb R$ but not uniformly on $\mathbb R$ (because $\sup u_n=1/2$ at $x=1/n$). On any closed interval $[a,b]$ with $a>0$ (away from origin), the convergence is uniform.

Step 5 — Standard reading + answer

Given the problem labels it as a “series of functions”, I’ll address the partial sums $S_N(x)=\sum_{n=1}^N u_n(x)$:

• $S_N(0)=0$, so converges at $x=0$.
• For $x\ne 0$, $S_N(x)\to\infty$ (since $u_n(x)\sim 1/(nx)$ has divergent sum) — series does not converge.

Hence the series diverges pointwise except at $x=0$, and does not converge uniformly on any set containing more than the single point $0$.

If the problem intended sequence (terms $u_n$):

Answer