← 2013 Paper 2

UPSC 2013 Maths Optional Paper 2 Q2c — Step-by-Step Solution

13 marks · Section A

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

Question

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

Technique

Leibniz/Dirichlet alternating-series test in the uniform form; comparison with harmonic series for absolute divergence.

Solution

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

Part 1 — Uniform convergence on $\mathbb{R}$

Strategy. Apply Leibniz’s alternating-series test in its uniform form. For an alternating series $\sum(-1)^{n-1}a_n(x)$ with $a_n(x)\ge 0$, monotone decreasing in $n$, and $a_n(x)\to 0$ uniformly in $x$, the series converges uniformly. Moreover, the truncation error after $N$ terms is bounded by $a_{N+1}(x)$.

Step 1.1 — $a_n(x)\ge 0$: $1/(n+x^{2})>0$ since $n\ge 1>0$ and $x^{2}\ge 0$ ✓.

Step 1.2 — Monotone decreasing in $n$: $a_{n+1}(x)=\dfrac{1}{n+1+x^{2}}<\dfrac{1}{n+x^{2}}=a_n(x)$ for each fixed $x$ ✓.

Step 1.3 — Uniform decay to zero:

The bound $1/n$ is independent of $x$, so $a_n(x)\to 0$ uniformly in $x\in\mathbb{R}$.

Step 1.4 — Conclude. By Leibniz’s test (uniform version), $\sum u_n(x)$ converges uniformly on $\mathbb{R}$. The truncation error after $N$ terms:

uniformly in $x$.

So the series is uniformly convergent on $\mathbb{R}$.

Part 2 — Not absolutely convergent (for any fixed $x$)

The absolute-value series is

Fix any $x\in\mathbb{R}$. For $n\ge x^{2}$, we have $n+x^{2}\le 2n$, so $\dfrac{1}{n+x^{2}}\ge\dfrac{1}{2n}$.

The tail $\displaystyle\sum_{n=\lceil x^{2}\rceil}^{\infty}\frac{1}{2n}$ diverges (harmonic-series tail). By the comparison test, $\displaystyle\sum\frac{1}{n+x^{2}}$ diverges.

So the series is not absolutely convergent for any real $x$.

Answer