Krdytkyu

I want to show that for sequence $(a_n)$, defined as $a_n=frac{6n+5}{2n+5}$, the following statement is true:

$$lim_{n rightarrow infty} {frac{6n+1}{2n+5}}=3$$

I started with stating that there exists an arbritary but fixed $ε>0$ and assumed that for an $n$ and $N_ε$, $n > N_ε$. I then used the Epsilon definition to prove that the limit is 3.

$$|a_n-3|<ε=|frac{6n+1}{2n+5}-3|<ε iff (frac{6n+1}{2n+5}-3)=frac{-7}{n}<ε iff frac{-7}{ε}=n$$

Here I set $frac{-7}{ε}=:N_ε$ and since we know that $n>N_ε$ that means that $$frac{1}{n}<frac{1}{ε}=frac{1}{frac{-7}{ε}}=frac{ε}{-7} ≠ ε$$

I don't understand where I went wrong in this proof, I've tried the same thing with other examples and get stuck in the same place.

You made one algebra error, and a number of other errors in the way your proof is organized and presented.

The algebra error, as José Carlos Santos notes, is that ${6n+1over2n+5}-3={-14over2n+5}$, not $-7over n$. So that of course affects what comes after. However, let's suppose $a_n-3={-7over n}$ is correct, and go from there.

The first, minor, problem is either a typo or a misuse of the equal sign: where you write

$$|a_n-3|ltepsilon=|{6n+1over2n+5}-3|ltepsilon$$

you really mean is $iff$ instead of $=$, i.e.,

$$|a_n-3|ltepsiloniff|{6n+1over2n+5}-3|ltepsilon$$

Also, at the other end of the display, where you write ${-7overepsilon}=n$, it should be ${-7overepsilon}lt n$.

The second, more serious, problem is that $|a_n-3|ltepsilon$ is not equivalent to $(a_n-3)ltepsilon$; the absolute value sign is important. So the first set of displayed expressions should end in

$$|a_n-3|=left|-7over nright|ltepsiloniff {7overepsilon}lt n$$

and thus we should set $N_epsilon={7overepsilon}$.

The next problem is merely, I think, a typo: where you say that $ngt N_epsilon$ means that ${1over n}lt{1overepsilon}$, you pretty clearly meant to write ${1over n}lt{1over N_epsilon}$, since you go on to substitute the expression you derived for $N_epsilon$ in the continuation.

The final, most significant, problem is one of logical relevance. You note, correctly, that $epsilonover-7$ (which, as noted above, shouldn't have the minus sign) is not equal to $epsilon$. But that doesn't matter. What you really need here is a set of implications along the lines of

$$ngt N_epsilonimplies{1over n}lt{1over N_epsilon}={epsilonover7}implies|a_n-3|={7over n}lt7cdot{epsilonover7}=epsilon$$

Of these, I'd say, the final problem is the most important: It looks like you were trying to derive the inequality ${1over n}ltepsilon$ instead of $|a_n-3|ltepsilon$. In other words, it looks like you lost track of what it is you are supposed to prove. It might help, for future proofs of this type, to keep in mind that, come hell or high water, you need to wrap things up with a logical sentence of the form $ngt N_epsilonimplies|a_n-L|ltepsilon$.

For some reason, your absolute value vanished. It turns ou that$$leftlvertfrac{6n+1}{2n+5}-3rightrvert=frac{14}{5+2n}.$$And$$frac{14}{5+2n}<varepsiloniff n>frac{14-5varepsilon}varepsilon.$$

Be careful:

When we're given an arbitrary $varepsilon>0$ we have to find $N_varepsilon$ such that, for $n>N_varepsilon$,
$$
left|frac{6n+1}{2n+5}-3right|<varepsilon
$$
We don't need to do “$iff$” as you seem to want. Now consider that
$$
left|frac{6n+1}{2n+5}-3right|=
left|frac{6n+1-3(2n+5)}{2n+5}right|=
left|frac{-14}{2n+5}right|=
frac{14}{2n+5}<frac{14}{2n}=frac{7}{n}
$$
We know have $7/n<varepsilon$ provided $n>7/varepsilon$, so we can take $N_varepsilon$ the largest integer less than or equal to $7/varepsilon$. For $n>N_varepsilon$ we thus have
$$
varepsilon>frac{7}{n}geleft|frac{6n+1}{2n+5}-3right|
$$
which ends our proof.

Be careful when you remove the absolute value sign and also how you look for upper bounds.