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I have the following question:

Find a sequence $(s_n)$ of real numbers such that $\lim_{n\to \infty} |s_n − s_{n+1}| = 0$, but $(s_n)$ does not converge.
However, prove that if the series $\sum_{n=0}^\infty|s_n−s_{n+1}|$ converges, then the sequence $(s_n)$ also converges.

My (very unsuccessful) work so far: If the series $\sum_{n=0}^\infty|s_n−s_{n+1}|$ converges, then the $\lim_{n\to \infty} |s_n − s_{n+1}| = 0$, so $s_n = s_{n+1} = s_{n+2} ...$ and hence $s_n$ is convergent. However based on the first part, this is clearly incorrect.
For the first part, I really can't think of anything at all, making it an alternating sequence is the only surefire way I can think of to assure $s_n$ does not converge.
Any help would be appreciated.

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If the series converges absolutely, then, with $s_0=0$, $$ s_{n}=\sum_{k=0}^{n-1} s_{k+1}-s_k, $$ so $$ \sum_{k=1}^\infty(s_{k+1}-s_k)=\lim_{n\to\infty}s_n. $$ As for the initial counterexample, let $$ s_n=\sum_{k=1}^n\frac1k. $$

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  • $\begingroup$ It is not a definition. It is an equality. $\endgroup$ – Martin Argerami Nov 8 '16 at 3:19
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EDIT

Note that this example does not give the answer to the problem. I wrote it to highlight the flawed reasoning at the end of the original post!

End Edit

Knowing that a limit goes to zero is insufficient to conclude that the individual terms go to zero. For example, let $s_n = \frac{1}{n}$. Then $$\lim_{n \to \infty} | s_n - s_{n+1} | = \lim_{n \to \infty} \frac{1}{n} - \frac{1}{n+1} = \lim_{n \to \infty} \frac{1}{n^2 + n} = 0$$

Note that it is never true that $s_n = s_{n+1}$. However, note that $\sum s_n$ is the familiar harmonic series, which diverges.

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  • $\begingroup$ Wow, that was actually really obvious, thanks for the help. Not sure how I didn't realize 1/n would work... $\endgroup$ – pretty fly for a pi guy Nov 8 '16 at 2:55
  • $\begingroup$ =) "Everything in mathematics is impossible until it's trivial" --Gromov (?). I struggled for a bit with trying to come up with a reasonable hint, but ended up caving. $\endgroup$ – erfink Nov 8 '16 at 3:06
  • $\begingroup$ @Dr.MV Edit added---I wasn't trying to solve the original problem, just point out the flaw in logic at the end of the OP, wherein it was basically concluded that all the $s_n$ were equal. $\endgroup$ – erfink Nov 8 '16 at 3:13
  • $\begingroup$ Right I knew it couldn't be that simple, I overlooked that when reading the answer by accident. $\endgroup$ – pretty fly for a pi guy Nov 8 '16 at 3:13

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