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I have the following problem. Prove by definition that the sequence

$r_{n}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...+\frac{(-1)^{n}}{2n+1}$

Is a Cauchy sequence.

I think I almost have it, but I have not been able to conclude anything of a series of inequalities that I obtained and I would like you to give me some advice or if I have an error or what I lack to conclude.

What I have is:

Sup. $n>m$

$|r_{n}-r_{m}|=|\sum_{k=m+1}^{n} \frac{(-1)^{k}}{2k+1}|\leq\sum_{k=m+1}^{n} \frac{1}{2k+1}<\sum_{k=m+1}^{n} \frac{1}{k}$

But from there I don't know what else to conclude

Haus
  • 734
  • Group the terms in $r_n-r_m$ two-by-two. – Kavi Rama Murthy Apr 29 '20 at 23:18
  • You should consider two consecutive terms in $\sum_{k = m+1}^{n} \dots$. Hence you'd have : $\frac{1}{2k+1} - \frac{1}{2k+2} \le \frac{1}{4k^2}$ – openspace Apr 29 '20 at 23:19
  • Looking at the sum of absolute values isn't going to help, because $\sum (1/k)$ is divergent, hence not Cauchy. –  Apr 29 '20 at 23:21
  • Thank you very much to all. – Haus Apr 29 '20 at 23:43
  • Instead of using the triangle inequality in the last equation, use the fact that alternating series (in the usual sense, where the absolute values of the terms are decreasing) are bounded in size by the absolute value of their first term. – Greg Martin Apr 29 '20 at 23:54

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