Question

Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N definition of convergence.

Answer 1

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$, there exists some $N$ such that whenever $n>N$, it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|<\varepsilon$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}<\varepsilon$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1<-\dfrac5{2\varepsilon}~\lor~2n-1>\dfrac5{2\varepsilon}$
$\implies n<\dfrac12-\dfrac5{4\varepsilon}~\lor~n>\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$, we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$.