1
$\begingroup$

If I want to determine whether a sequence, ${a_n}$, is bounded above $\forall n \in \Bbb{N} $, is it enough to find a sequence that is larger than $a_n$, and show that it converges and is therefore bounded? For example:

$\forall n \in \Bbb{N}, let,$

$$ a_n = \frac{1}{n+1} + \frac{1}{n+2} + ...+\frac{1}{2n}\\ \frac{1}{n+1} + \frac{1}{n+2} + ...+\frac{1}{2n} \leq \frac{1}{n} + \frac{1}{n} + ...+\frac{1}{n} = n\cdot\frac{1}{n}=1 $$ and since $\lim\limits_{n\to\infty}1 = 1$, then 1 must be an upper bound for $a_n$. Is this correct? Thanks.

$\endgroup$
  • $\begingroup$ You have shown that $a_n \le 1$, no convergence or limit is needed here. $\endgroup$ – Martin R Aug 15 '18 at 9:13
  • $\begingroup$ Incidentally you have for large $n$: $a_n $ $=H_{2n}-H_n$ $\approx (\log_e(2n) +\gamma +\frac{1}{4n})- (\log_e(n)+\gamma+\frac{1}{2n})$ $=\log_e(2) - \frac{1}{4n}$ $\approx 0.693-\frac{1}{4n}$ $\endgroup$ – Henry Aug 15 '18 at 9:17
2
$\begingroup$

I think you mess up some ideas.

You say "and since $\lim\limits_{n\to\infty}1 = 1$", but you never showed that $\lim\limits_{n\to\infty}1 = 1$. And if you check the comment of Henry this seems to be wrong. But you don't need the limes.

You showed that

$$a_n = \frac{1}{n+1} + \frac{1}{n+2} + ...+\frac{1}{2n}\\ \frac{1}{n+1} + \frac{1}{n+2} + ...+\frac{1}{2n} \leq \frac{1}{n} + \frac{1}{n} + ...+\frac{1}{n} = n\cdot\frac{1}{n}=1$$ this means

$$a_n\le 1,\; \forall n \in \Bbb{N}$$

And this means that $a_n$ is bounded above by $1$. There is nothing else to show.

Remark 1: An increasing sequence that is bounded above is convergent

We have $$a_{n+1}=a_n+\frac{1}{(2n+1)(2n+2)}$$ This means $$a_{n+1}>a_n$$ and so $a_n$ is monotone increasing. If a sequence is increasing and bounded above then the sequence is convergent.

Remark 2: An convergent sequence is bounded

If a sequence $a_n$ converges to $a$ then there exists a number $N$ such that $$\mid a_n-a\mid\le 1,\; \forall n>N$$ and so we have $$a_n\le a+1,\; \forall n>N$$ and $$a_n\le\max\{a_1,\ldots,a_N\},\; \forall n\le N$$ and therefore the sequence $a_n$ is bounded by $$\max\{N,a_1,\ldots,a_N\}$$

$\endgroup$
  • $\begingroup$ Thanks very much for this! The question I was tackling actually asked to show whether the sequence converges or not, and I had already shown that it was increasing, so it's good to get confirmation of that. The class I'm taking is an introduction to real analysis, and it's taking some getting used to! I appreciate the remarks, very helpful. $\endgroup$ – monkeyofscience Aug 15 '18 at 10:54
0
$\begingroup$

Yes, it is correct, we have $a_n \le 1$ for all $n$.

$\endgroup$
0
$\begingroup$

Yes, this is enough. If $(b_n)$ dominates $(a_n)$ and $b_n \rightarrow b$, then there is $S\in \mathbb R$ with $b_n<S$ for all $n \in \mathbb N$. Since $(b_n)$ dominates, we get $S>b_n \geq a_n$, so $(a_n)$ is bounded as well.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.