Prove that the series $\sum_{n=1}^{\infty} a_n$ converges absolutely if and only if for every $\epsilon > 0$, there exists a positive integer $N$ such that $$|a_{n_1}+a_{n_2}+\ldots+a_{n_k}|<\epsilon$$ wherever $n_1,\ldots,n_k$ are distinct positive integers exceeding $N$.

Forward direction: Suppose the series $\sum_{n=1}^{\infty} a_n$ converges absolutely. That means the sequence $|a_1|,|a_1|+|a_2|,\ldots,|a_1|+|a_2|+\ldots+|a_n|, \ldots$ converges, so it is Cauchy, so for every $\epsilon$ there exists $N$ such that $|a_p|+|a_{p+1}|+\ldots+|a_q|<\epsilon$ for all $N<p<q$. Then $|a_{n_1}+a_{n_2}+\ldots+a_{n_k}| < |a_{n_1}|+|a_{n_2}|+\ldots+|a_{n_k}|\leq |a_{n_1}|+|a_{n_1+1}|+\ldots+|a_{n_k}|<\epsilon$.

What about the backward direction?

  • $\begingroup$ You seem to have a confusion about what absolute convergence means. $\endgroup$ – Mariano Suárez-Álvarez Jun 12 '13 at 4:44
  • $\begingroup$ Oops, I typed too quickly. Edited. $\endgroup$ – Paul S. Jun 12 '13 at 4:47
  • $\begingroup$ There you go :-) $\endgroup$ – Mariano Suárez-Álvarez Jun 12 '13 at 4:50
  • $\begingroup$ If the sum doesn't converge absolutely then for any $N$, $\Sigma _{n>N} |a_n| = \Sigma _{n>N, a_n >0} a_n + \Sigma _{n>N, a_n <0} -a_n =\infty$. So at least one of those two series has to diverge... $\endgroup$ – Tim kinsella Jun 12 '13 at 5:06

By contradiction, fix any $\varepsilon>0.$ If the series does not converge absolutely, then for each $N,$ there is $M(N)>N$ such that $$|a_N|+|a_{N+1}|+...|a_{M(N)}|>2\varepsilon.$$ Now, let $a_{m_N},a_{m_{N+1}},...a_{m_k}\ge 0$ be nonnegative numbers among $a_N,...a_{M(N)}$ and the rest $a_{m_{k+1}},a_{m_2},...a_{m_{M(N)}}<0.$ Then either $$|a_N|+|a_{N+1}|+...|a_{M(N)}|=|a_{m_N}+a_{m_{N+1}}+...+a_{m_k}|+|a_{m_{k+1}}+a_{m_2}+...+a_{m_{M(N)}}|$$ or $$|a_N|+|a_{N+1}|+...|a_{M(N)}|=|a_{m_{k+1}}+a_{m_2}+...+a_{m_{M(N)}}|+|a_{m_N}+a_{m_{N+1}}+...+a_{m_k}|$$ In both cases, either $|a_{m_{k+1}}+a_{m_2}+...+a_{m_{M(N)}}|>\varepsilon$ or $|a_{m_N}+a_{m_{N+1}}+...+a_{m_k}|>\varepsilon.$ Since $N$ was chosen arbitrary, we arrive at contradiction.

  • $\begingroup$ Actually, leshik, isn't it true that $|a_N|+|a_{N+1}|+...|a_{M(N)}|=|a_{m_N}+a_{m_{N+1}}+...+a_{m_k}|+|a_{m_{k+1}}+a_{m_2}+...+a_{m_{M(N)}}|$ (The sign in the middle should be plus, not minus) $\endgroup$ – Paul S. Jun 12 '13 at 19:38
  • $\begingroup$ You can probably fix it by using $\varepsilon/2$ at the end instead of $\varepsilon$. $\endgroup$ – Paul S. Jun 12 '13 at 19:42
  • $\begingroup$ thanks, I have edited the signs. The proof does not change. $\endgroup$ – leshik Jun 12 '13 at 19:42
  • $\begingroup$ Oh, so you use $2\varepsilon$ at the beginning instead. Thanks! $\endgroup$ – Paul S. Jun 12 '13 at 19:43

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