2
$\begingroup$

If $\{b_n\}$ is a sequence of nonnegative numbers that converges to $0$, and $\{x_n\}$ is a real sequence that satisfies $|x_n-a|\le Cb_n$ for large $n$, where $C$ is a fixed positive constant, prove that $x_n$ converges to $a$.

I understand this is a simple question. I feel that I am missing a very simple detail because my attempt isn't consistent with the hint that is given (or, I just don't understand what I'm doing). I realize problems have many different ways they can be solved. I am new to sequences and I am unsure if what I did makes sense or not.

The hint provided is that any $\epsilon$, including $\large\frac{\epsilon}{c}$, when, $c\gt 0$, works for the definition: A sequence of real numbers $\{x_n\}$ is said to converge to a real number $a\in \mathbb R$ if and only if for every $\epsilon\gt 0$ there is an $N\in \mathbb N$ such that $n\ge N \to |x_n-a|\lt \epsilon$.

My Attempt:

If $b_n$ converges to $0$, then $|b_n|\to 0$ as $n\to \infty$

$|b_n|\to 0$
$C \cdot|b_n|\to 0\cdot C$
$C|b_n|\to 0$

$|x_n-a|\le Cb_n$
Since, $Cb_n\to 0$, then $|x_n-a|\to 0$.
By definition, if $|x_n-a|\to 0$ as $n\to \infty$, then $x_n$ converges to $a$.

Any input would be great.

$\endgroup$
1
  • $\begingroup$ I agree with the answer given that they are likely anticipating it being done via the definitions judging from the hint; however your way works too, but would be better with more words to link the steps together. Here's what I'd go for: Suppose $(b_n) \rightarrow 0$. Then, $(C b_n) \rightarrow 0$ by the sum rule. As we have $$ 0 \leq \vert x_n - a \vert \leq C b_n, $$ by the sandwich rule, $$\vert x_n - a \vert \rightarrow 0,$$ hence $(x_n) \rightarrow a$. $\endgroup$
    – Matt Rigby
    Sep 12, 2014 at 7:32

1 Answer 1

2
$\begingroup$

The idea is good but you need to be more rigorous and stick to the definition:

Let $\epsilon > 0$, then $$\lim_{n \to \infty}b_n = 0$$ implies that there is some $N \in \mathbb{N}$ such that $$b_n=|b_n|<\frac{\epsilon}{ C}$$ for every $n >N$. It follows that $$|x_n-a|\leq Cb_n < C\frac{\epsilon}{ C} = \epsilon$$ for every $n>N$, i.e. $$\lim_{n \to \infty}x_n = a.$$

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .