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.

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



when is arbitrarily large.

Now, for the limit as to be equal to is to say that for any , there exists some such that whenever , it follows that



From this inequality, we get






As we're considering , we can omit the first inequality.

We can then see that choosing will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that .

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