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In my textbook, it states the following theorem:

If $\lim\limits_{n \to\infty} |a_n|=0$ then $\lim\limits_{n \to\infty} a_n=0$

My question is, if I find that the limit of the absolute value of the sequence does not equal $0$, can I conclude that the limit of the actual sequence then diverges? Or does the theorem only work one way?

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3 Answers 3

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If $|a_n|$ does not converge to $0$, then you can certainly say that neither $a_n$ converges to $0$. However, if $|a_n|$ does not converge to $0$, you cannot say that $(a_n)$ diverge. Take for instance as $(a_n)$ the constant sequence equal to 1.

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  • $\begingroup$ So, If the absolute value of my sequence tends to any number other than 0, or even infinity, I cannot conclude on the convergence or divergence of my original sequence? Unless it tends to 0? $\endgroup$
    – melm
    Nov 26, 2016 at 17:00
  • $\begingroup$ Exactly. Infact, let $l$ be a real number and set $a_n = (-1)^n l$ that $|a_n| =l$ which converges to zero, but $(a_n)$ does not converge... $\endgroup$
    – Marsan
    Nov 26, 2016 at 17:13
  • $\begingroup$ Why does |an|=l converge to 0? $\endgroup$
    – melm
    Nov 26, 2016 at 17:27
  • $\begingroup$ if $l \neq 0$ of course $\endgroup$
    – Marsan
    Nov 26, 2016 at 17:48
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Here are two examples: $$ 2.1,\ 2.01,\ 2.001,\ 2.0001,\ 2.00001, \ \ldots\ldots $$ $$ 2,\ {-2},\ 2,\ {-2},\ 2,\ {-2}, 2,\ {-2},\ 2,\ {-2},\ \ldots $$

In both cases, the absolute value converges to $2$, and $2$ is not $0$.

The first sequence converges.

The second sequence does not converge.

However, if the sequence of absolute values diverges to $+\infty$, then in every case the original sequence (without absolute values) fails to converge.

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  • $\begingroup$ Dang, almost my answer in spirit. :D $\endgroup$ Nov 26, 2016 at 17:18
  • $\begingroup$ sorry just to conclude, if the limit of the absolute tends to 0, the original tends to 0 and converges. If the limit of my absolute diverges, then I can conclude that the original sequence also diverges in all cases? But other than that no conclusion can be made for a sequence tending to any other real number? Have I got it right? $\endgroup$
    – melm
    Nov 26, 2016 at 17:35
  • $\begingroup$ @aa21 : The absolute value function is continuous; therefore if $a_n$ converges to something as $n\to\infty$ then so does $|a_n|$; therefore if $|a_n|$ fails to converge then so does $|a_n|$. But if $a_n$ fails to converge then $|a_n|$ may converge or fail to converge; however, if $a_n$ fails to converge, then $a_n$ converges only if $a_n$ has only two subsequential limits, both having the same absolute value. $\qquad$ $\endgroup$ Nov 26, 2016 at 19:34
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It is a one way statement. Consider the following sequences:

$$\{+1,-1,+1,-1,\dots\}$$

$$\left\{\frac21,\ \frac32,\ \frac43,\ \frac54,\dots\right\}$$

Then both have $\lim_{n\to\infty}|a_n|=1$, but one diverges while the other converges.


However, notice that if $\lim_{n\to\infty}|a_n|=c$, then the sequence is bounded.

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