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? 
 A: 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.
A: 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.
