← 2017 Paper 2

UPSC 2017 Maths Optional Paper 2 Q4c — Step-by-Step Solution

20 marks · Section A

Rearrangement of Series; Riemann's Theorem · Real Analysis · Read the full method →

Question

Let $\displaystyle\sum_{n=1}^\infty x_n$ be a conditionally convergent series of real numbers. Show that there is a rearrangement $\displaystyle\sum_{n=1}^\infty x_{\pi(n)}$ of the series $\displaystyle\sum_{n=1}^\infty x_n$ that converges to $100$.

Technique

Riemann rearrangement — split into positive/negative subseries (both divergent, terms $\to0$, from conditional convergence), greedily overshoot/undershoot the target $100$, bound each crossing error by a single term $\to0$.

Solution

This is a special case (target $S=100$) of the Riemann rearrangement theorem. The proof has three parts: (A) extract the positive and negative subsequences and establish their key properties; (B) describe a greedy rearrangement; (C) prove it converges to $100$.

Step 1 — Positive and negative parts; consequences of conditional convergence

Let $p_n=\max(x_n,0)\ge0$ (positive parts) and $q_n=\max(-x_n,0)\ge0$ (negative parts), so

Conditional convergence means $\sum x_n$ converges but $\sum|x_n|=\infty$.

Claim: $\displaystyle\sum p_n=+\infty$ and $\displaystyle\sum q_n=+\infty$.

Suppose, for contradiction, that $\sum p_n<\infty$. Since $q_n=p_n-x_n$ and both $\sum p_n$ and $\sum x_n$ converge, $\sum q_n=\sum p_n-\sum x_n$ would converge; then $\sum|x_n|=\sum(p_n+q_n)<\infty$, contradicting conditional convergence. Symmetrically $\sum q_n<\infty$ also fails. Hence both $\sum p_n=\sum q_n=+\infty$.

Also: $x_n\to0$ (terms of a convergent series), so $p_n\to0$ and $q_n\to0$.

Now discard the zero terms and list the strictly positive terms of the series in their original order as $a_1,a_2,a_3,\ldots$ (with $\sum a_k=+\infty$, $a_k\to0$) and the absolute values of the strictly negative terms as $b_1,b_2,b_3,\ldots$ (with $\sum b_k=+\infty$, $b_k\to0$). Each original nonzero $x_n$ appears exactly once among the $a$‘s or the $b$‘s; this is what lets us build a genuine rearrangement (a bijection $\pi$). (Zero terms, if any, can be inserted anywhere and do not affect convergence; place them, say, after each negative block.)

Step 2 — The greedy construction

Build the rearranged series in blocks, alternating “add positives until you exceed $100$” and “subtract positives’ counterpart, i.e. add negatives until you drop below $100$”:

1. Take the fewest leading positive terms $a_1,\ldots,a_{m_1}$ so that the partial sum first exceeds $100$:

(Possible because $\sum a_k=+\infty$, so the positive sums are unbounded.) 2. Then append the fewest leading negative terms $-b_1,\ldots,-b_{n_1}$ so the running sum first drops below $100$:

with the previous step ($n_1-1$ negatives) still $\ge100$. (Possible because $\sum b_k=+\infty$.) 3. Resume with the next unused positives $a_{m_1+1},\ldots,a_{m_2}$ until the sum again first exceeds $100$; then the next unused negatives until it again first drops below $100$; and so on indefinitely.

Because both $\sum a_k$ and $\sum b_k$ diverge, each step can always be completed, and every $a_k$ and $b_k$ is eventually used (we always continue from where we left off and each block uses at least one new term and ratchets through the lists). So this defines a bijection $\pi$ of $\mathbb N$, i.e. a genuine rearrangement $\sum x_{\pi(n)}$.

Step 3 — Convergence to $100$

Let $S_N$ denote the partial sums of the rearranged series. By construction the running sum crosses the level $100$ at the end of each block, alternately overshooting above and undershooting below. Consider any “turning point” partial sum:

• At the end of a positive block (sum just $>100$): the overshoot is caused by adding the single last positive term $a_{m_j}$, and just before adding it the sum was $\le100$. Hence

• At the end of a negative block (sum just $<100$): similarly the undershoot is bounded by the single last negative term $b_{n_j}$:

Between consecutive turning points the partial sums are monotone (adding only positives, or only negatives), so every partial sum $S_N$ lies within distance $\max(a_{m_j},b_{n_j})$ of $100$ for the current block. Precisely, for all $N$ in the $j$th pair of blocks,

As $j\to\infty$ we use infinitely many terms, so the indices $m_j,n_j\to\infty$; since $a_k\to0$ and $b_k\to0$, we get $a_{m_j}\to0$ and $b_{n_j}\to0$. Therefore

i.e. $S_N\to100$. Hence

Answer