Prove that $\{s_n\}$ convergence implies $\{\left|s_n\right|\}$ convergence?

Proof. Let $\{s_n\}$ be a sequence of real numbers. We show that $s_n \to s$ implies $|s_n| \to t$ for $s, t \in \Bbb{R}$. By the Cauchy criterion then $\exists p, q \in \Bbb{N}$ s.t. $\left|\sum_{n=p}^q s_n\right| < \epsilon$.

I don't know how to continue.

I need to show $$\left|\sum_{n=p}^q \left|s_n\right|\right| < \epsilon$$

But that latter sequence is larger than the former...so I can't use comparison test.

This is from Rudin, Chapter 3 exercise 1.

  • 4
    $\begingroup$ You seem to be confusing Cauchy criterion applied to $\sum\limits_ns_n$ and Cauchy criterion applied to $s_n$. $\endgroup$ – Did Dec 3 '13 at 21:52
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    $\begingroup$ Don't try to shove a test in, rather look directly at what's going on. Note that there are three cases, according to whether $s_n\to0$, $s_n\to\alpha>0$ or $s_n\to\beta<0$. You should be able to prove easily that $|s_n|$ converges in the last two cases. For the first, just look at the formal definition of "$s_n\to 0$". $\endgroup$ – Andrés E. Caicedo Dec 3 '13 at 21:56
  • $\begingroup$ Hint: Use the triangle inequality. $\endgroup$ – Christopher A. Wong Dec 3 '13 at 22:04
  • $\begingroup$ @Andres: I'll give it a shot and report back. Don't leave me hanging! $\endgroup$ – Don Larynx Dec 3 '13 at 22:04
  • $\begingroup$ See here for a proof. $\endgroup$ – jdoicj Jul 1 '14 at 7:12

Hmmm... With one case and only one case: assume that $s_n\to\ell$, that is, that $|s_n-\ell|\to0$. Then $||s_n|-|\ell||\leqslant|s_n-\ell|$ for every $n$ hence $||s_n|-|\ell||\to0$, that is, $|s_n|\to|\ell|$.

Keyword: Second triangular inequality.

  • $\begingroup$ But if $|s_n|$ converges means it satisfies cauchy criterion so $| |s_m|-|s_n| | = | |x_{m-1}|+|x_{m-2}|+ \dots |x_{n+1}| | \leq \epsilon $ i.e $|x_{m-1}+x_{m-2}+ \dots x_{n+1} | \leq |x_{m-1}|+|x_{m-2}|+ \dots |x_{n+1}| \leq \epsilon$ So $s_n$ converges. What is wrong in this ? $\endgroup$ – Rising Star Nov 10 '15 at 13:12
  • $\begingroup$ @RisingStar Simply that $||s_m|-|s_n||=||x_{n+1}|+\ldots+|x_{m}||$ is not true in general. You might want to compute every $s_n$ when $x_1=+1$ and $x_n=2(-1)^{n+1}$ for every $n\geqslant2$, to check that $(|s_n|)$ converges but not $(s_n)$. $\endgroup$ – Did Nov 10 '15 at 14:55

Suppose $(s_n)$ is a sequence of real numbers and $s_n\to s$. There are three cases, depending on whether $s>0$, $s<0$, or $s=0$.

Suppose first that $s\ne0$. Recall that $s_n\to s$ means that for any $\epsilon>0$ there is an $N$ such that if $n>N$ then $|s_n-s|<\epsilon$. Specifically, let $\epsilon=|s|$ (or any positive number less than $|s|$). Since $|s|=|s-0|$, the condition $|s_n-s|<|s|$ (valid for all $n>N$) means that $s_n$ is closer to $s$ than $s$ is to $0$, so $s_n$ and $s$ must have the same sign. In the first case, where $s>0$, this means that $|s_n|=s_n$ for all $n>N$, so $|s_n|\to s$. In the second case, where $s<0$, we have $|s_n|=-s_n$ for all $n>N$, so $|s_n|\to -s$.

We are left with the case where $s=0$. Again, that $s_n\to 0$ means that for any $\epsilon>0$ there is an $N$ such that for all $n>N$, we have $|s_n-0|<\epsilon$. But $$ |s_n-0|=|s_n|=|s_n|-0=||s_n|-0|, $$ and we have shown that for all $\epsilon>0$ there is an $N$ such that if $n>N$ then $||s_n|-0|<\epsilon$, that is, $|s_n|\to 0$.

Finally, note that in all cases we have that $|s_n|\to|s|$. What we have shown is a particular instance of a general fact, namely, that if $f$ is continuous at a point $s$, $(s_n)$ is a sequence of points such that $s_n\to s$, and all the $s_n$ are in the domain of $f$, then $f(s_n)\to f(s)$. In this case, $f(x)=|x|$. Actually, in the setting we are discussing, the converse holds, so continuity of $f$ at $s$ is equivalent to this fact, that is, if $s$ is in the domain of $f$, then $f$ is continuous at $s$ iff whenever $s_n\to s$ and all the $s_n$ are in the domain of $f$, then $f(s_n)\to f(s)$. So we can take the argument above as a (convoluted) proof that $f(x)=|x|$ is continuous.


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