Prove from definition of convergence that (-2n+5)/(3n+1) is convergent.

Question

Prove directly from the definition of convergence that $\frac{-2n+5}{3n+1}$ is convergent.

So I let $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $n \geq to N$ implies that the $\left | \frac{-2n+5}{3n+1}-0 \right | \leq \frac{-2n+5n}{3n} = 1 \leq \epsilon $

But, I am looking to set $N$ equal to some number, so $1 \leq \epsilon$ isn't an ideal solution! Can anyone please help?

Answer 1

Let $\epsilon>0$. We have

$$\left|\frac{-2n+5}{3n+1}+\frac23\right|=\frac{17}{9n+3}<\epsilon\iff n>\frac19\left(\frac{17}{\epsilon}-3\right)=:\alpha$$ so for $n_0=\max(0,\lfloor\alpha\rfloor+1)$ we have for $n\ge n_0$ $$\left|\frac{-2n+5}{3n+1}+\frac23\right|<\epsilon$$ hence we proved by definition that the limit is $-\frac23$.

Answer 2

Hint: As Yves points out above, the limit of this sequence is $-\frac{2}{3}$. So given $\epsilon$ you want to find $N>0$ such that whenever $n>N$ we have \begin{equation*} \left|\frac{-2n+5}{3n+1}-\frac{-2}{3}\right| < \epsilon. \end{equation*} Now, \begin{equation*} \frac{-2n+5}{3n+1}-\frac{-2}{3} = \frac{17}{3(1+3n)}. \end{equation*} So now given $\epsilon$, can you figure out what $N$ has to be?

Answer 3

Let $\epsilon>0$ be given. Then

$\displaystyle \left|\frac{-2n+5}{3n+1}-\big(-\frac{2}{3}\big)\right|=\left|\frac{17}{3(3n+1)}\right|=\frac{17}{3(3n+1)}<\frac{18}{9n}=\frac{2}{n}$, so

let $ N\in\mathbb{N}$ with $\displaystyle N>\frac{2}{\epsilon}$.

If $n\ge N$, then $\displaystyle n>\frac{2}{\epsilon}\implies\frac{2}{n}<\epsilon\implies\left|\frac{-2n+5}{3n+1}-\big(-\frac{2}{3}\big)\right|<\epsilon$.