While practicing for an exam, I encountered this question: (e) For a sequence $\lbrace b_n \rbrace_{n=1}^\infty \subset \mathbb{R}$ is given: $\forall n \in \mathbb{N}$, $\: b_n < b_{n+1} < 2$.

Does $\lim_{n\to\infty} b_n$ exist? If $\lim_{n\to\infty}$ exists, give $\lim_{n\to\infty}$and explain the answer. If $\lim_{n\to\infty}$ does not exist, explain the answer.

My thought process: The limit exists $\Longleftrightarrow $ if $\exists n$ such that $b_n=b_{n+1}$. And because this contradicts with the fact that $b_n < b_{n+1}$. Hence, the limit does not exist.

A classmate on mine had an other answer: "Since $b_{n+1}>b_n$ but $b_{n+1} < 2$, we know that $b_{n+1}$ is closer to 2 than $b_n$. He deduces from this that the limit is 2.

Who of us is right?

  • $\begingroup$ Both of you are wrong: the limit exists, but it may not be equal to $2$. $\endgroup$ – njguliyev Oct 28 '13 at 22:21
  • $\begingroup$ For your claim, the sequence $b_n = 2-\frac{1}{n}$ satisfies $b_n\neq b_{n+1},\,\forall n$, but $\lim_{n\rightarrow\infty}b_n = 2$. $\endgroup$ – Lord Soth Oct 28 '13 at 22:22
  • $\begingroup$ In a similar manner, for your friend's claim, think about the sequence $1.9876-\frac{1}{n}$. $\endgroup$ – Lord Soth Oct 28 '13 at 22:23
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    $\begingroup$ hint: consider $\beta = \sup\{(b_n): n \in \mathbb{N}\}$ $\endgroup$ – B. Mackey Oct 28 '13 at 22:23
  • $\begingroup$ ... and $b_n=1-\frac1n$ satisfies the condition of the problem as well, its limit exists but is $\ne 2$. $\endgroup$ – Hagen von Eitzen Oct 28 '13 at 22:23

Your sequence is increasing and bounded. Thus it is convergent.

The limit $l$ satisfies $l \leq 2$, but as it was pointed it is not necessarily $2$. For each $ l\leq 2$ $$b_n =l -\frac{1}{n}$$

is an example of a sequence satisfying the conditions and converging to $l$.


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