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
`\frac32` because


`(3n+1)/(2n-1)=(3+\frac1n)/(2-\frac1n)\approx\frac32`

when
`n` is arbitrarily large.

Now, for the limit as
`n\to\infty` to be equal to
`\frac32` is to say that for any
`\varepsilon>0`, there exists some
`N` such that whenever
`n>N`, it follows that


`\left|(3n+1)/(2n-1)-\frac32\right|<\varepsilon`

From this inequality, we get


`\left|(3n+1)/(2n-1)-\frac32\right|=\left|((6n+2)-(6n-3))/(2(2n-1))\right|=\frac52\frac1<\varepsilon`

`\implies|2n-1|>\frac5{2\varepsilon}`

`\implies2n-1<-\frac5{2\varepsilon}~\lor~2n-1>\frac5{2\varepsilon}`

`\implies n<\frac12-\frac5{4\varepsilon}~\lor~n>\frac12+\frac5{4\varepsilon}`

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

We can then see that choosing
`N=\left\lceil\frac12+\frac5{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`.