IMC 1997 · Day 1 · P2

a) Yes. Let $S = \sum\limits_{n=1}^{\infty} a_n$, $S_n = \sum\limits_{k=1}^{n} a_k$. Fix $\varepsilon > 0$ and a number $n_0$ such that $|S_n - S| < \varepsilon$ for $n > n_0$. The partial sums of the permuted series have the form $L_{2^{n-1}+k} = S_{2^{n-1}} + S_{2^n} - S_{2^n - k}$, $0 \le k < 2^{n-1}$ and for $2^{n-1} > n_0$ we have $|L_{2^{n-1}+k} - S| < 3\varepsilon$, i.e. the permuted series converges.

b) No. Take $a_n = \dfrac{(-1)^{n+1}}{\sqrt{n}}$. Then $L_{3 \cdot 2^{n-2}} = S_{2^{n-1}} + \sum\limits_{k=2^{n-2}}^{2^{n-1}-1} \dfrac{1}{\sqrt{2k+1}}$ and $L_{3 \cdot 2^{n-2}} - S_{2^{n-1}} \ge \dfrac{2^{n-2}}{\sqrt{2^n}} \xrightarrow[n \to \infty]{} \infty$, so $L_{3 \cdot 2^{n-2}} \xrightarrow[n \to \infty]{} \infty$.