# Proof to sequences in real analysis

I need some verification for my proof in part a) and help to get me started on part b)

a) Prove that the sequence $a_n = (2n+1)/(3n+5)$ converges to $2/3$ directly from the definition of convergence of a sequence.

Solution: Some rough work first: Given $\epsilon>0$, we want $|a_n-(2/3)|< \epsilon$. Need $|(2n+1)/(3n+5) – (2/3)|< \epsilon$ which after simplifying gives $7/(9n+15)< \epsilon$. Solving for $n$ gives $n> (7-15\epsilon)/9\epsilon$.

Then we write our formal proof: Given $\epsilon>0$, set $N= (7-15\epsilon)/9\epsilon$. Let $n>N$. Then $|a_n-(2/3)| = 7/(9n+15) < 7/(9((7-15\epsilon)/9\epsilon)+15) =\epsilon$.
Thus by definition the sequence converges to 2/3.

b) Suppose that the sequence $\{a_n\}$ is not bounded above (that is, for any real number $x$, there is some $n$ so that $a_n>x$). Prove that $\{a_n\}$ does not converge to $42$.

To solve this I am assuming that we have to suppose that $\{a_n\}$ does converge to $42$, and then we will get a contradiction, that is the sequence will be bounded above. So can anyone help me get started on this?

#### Solutions Collecting From Web of "Proof to sequences in real analysis"

Your answer to part (a) is well thought through. However your post for part (b) has confused the assumption you need for a proof.

For part (b): If $\{a_n\}$ is not bounded above, suppose for a contradiction that $\{a_n\}$ converges to $42$. Let us fix $\varepsilon=1$ and choose $N\in\mathbb{N}$ such that for every $n>N$, $|a_n-42|<1$. (All we have done is use the definition of convergence.) Since

$$|a_n-42|<1 \iff -1<a_n-42<1,$$

whenever $n>N$ we have that $a_n<43$. Now let us concentrate on a global upper bound, one that includes the finite number of terms $a_1, a_2, \ldots ,a_{N-1}, a_{N}$ (using the same value of $N$ as before).

The finite number of terms are bounded above by $\max_{1\le i\le N} a_i$. It follows that for every $n\in\mathbb{N}$ we have the following bound:

$$a_n\le \max\left\{ \max_{1\le i\le N}a_i, 43\right\}.$$ Clearly $\{a_n\}$ is bounded above, a contradiction.

Hint for (b): if a sequence converges to $42$ (or any other number), at most a finite number of terms fail to belong to a certain neighborhood of $42$.

For (a) the proof is essentially correct, but you have to take $N$ as an integer greater than or equal to $(7-15\varepsilon)/(9\varepsilon)$, which exists by Archimedes’ property.