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If $\{x_n\}$ is a sequence of positive real numbers which is not bounded, then it diverges to infinity. State whether the above statement is true or false. If true/false , give the reason.

I just know that if the sequence is of positive real numbers then it must be either increasing or decreasing sequence or it would be constant sequence.

My question is what can I conclude about the word "not bounded"? Does it means not bounded above or not bounded below or neither bounded above nor bounded below? Anyone just help me to draw conclusion whether the statement is true or not?

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  • $\begingroup$ No neither bounded above,nor below? What does it mean? $\endgroup$
    – Kavita
    Oct 5, 2015 at 11:53
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    $\begingroup$ You are given a sequence of POSITIVE real numbers. So it is obviously bounded below by zero. If they talk about it being unbounded the only possibility here is unbounded above. As such either unbounded below or above qualifies it to be unbounded. $\endgroup$
    – R_D
    Oct 5, 2015 at 11:55
  • $\begingroup$ I agree with you,thank you. $\endgroup$
    – Kavita
    Oct 5, 2015 at 11:56
  • $\begingroup$ You should also observe that there are things called oscillating sequences. Not all sequences of real numbers are increasing, decreasing, or constant $\endgroup$
    – R_D
    Oct 5, 2015 at 11:58
  • $\begingroup$ Uups!! Yes there are things called oscillating sequences. So the statement is false? Can i conclude this? $\endgroup$
    – Kavita
    Oct 5, 2015 at 12:03

3 Answers 3

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A sequence $(x_{n})$ of reals is bounded iff there is some $M \geq 0$ such that $|x_{n}| \leq M$ for all $n\geq 1$; the sequence $(x_{n})$ diverges to infinity iff for every $B > 0$ we have $|x_{n}| \geq B$ eventually. Now if $(x_{n})$ is not bounded and if $x_{n} > 0$ for all $n \geq 1$, then for every $M > 0$ there is some $n \geq 1$ such that $x_{n} > M$; so $(x_{n})$ need not diverge to infinity by definition. For instance, the sequence $(x_{n})$ defined by $x_{n} := n + (-1)^{n-1}(n-1)$ for all $n \geq 1$ is unbounded and does not diverge to infinity.

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  • $\begingroup$ That is not the definition for a diverging sequence. $\endgroup$
    – Muschkopp
    Oct 5, 2015 at 12:11
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    $\begingroup$ @Muschkopp: Yes it is not; I should have and added "to infinity". Thank you for good eyes. $\endgroup$
    – Yes
    Oct 5, 2015 at 12:18
  • $\begingroup$ yes. the statement is indeed true if we replace "diverges to infinity" with simply "diverges". This is simply the contrapositive of convergent => bounded $\endgroup$
    – Paul
    Aug 19, 2018 at 0:23
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The statement is false. For example the sequence 1,2,1,4,1,5,1,6,1,7,... is unbounded sequence of positive real numbers that does not diverge to infinity. It is oscillating sequence.

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I just know that if the sequence is of positive real numbers then it must be either increasing or decreasing sequence or it would be constant sequence.

Then what can you say about the sequence defined by

$$a_n = n(1+(-1)^n)$$

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  • $\begingroup$ This is not a sequence of positive real numbers. $\endgroup$
    – Muschkopp
    Oct 5, 2015 at 12:06
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    $\begingroup$ @Muschkopp Just consider $1+a_n$ then. $\endgroup$ Oct 5, 2015 at 12:07
  • $\begingroup$ @Najib Idrissi Of course. Just wanted to point that out. $\endgroup$
    – Muschkopp
    Oct 5, 2015 at 12:09
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    $\begingroup$ @Muschkopp : sorry, I always forget that positive in english means strictly greater than 0. $\endgroup$
    – Tryss
    Oct 5, 2015 at 12:17

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