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myrzilka [38]
3 years ago
11

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

tion of convergence.
Mathematics
1 answer:
Nat2105 [25]3 years ago
6 0
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|

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|}
\implies|2n-1|>\dfrac5{2\varepsilon}
\implies2n-1\dfrac5{2\varepsilon}
\implies 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.
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