We want to show that the sequence \(\left\{ 2 + \frac{1}{n} \right\}\) approaches a limit of \(2\).
In order to show this, we must show that any open interval containing \(2\), no matter how small, must contain a tail of the sequence.
Notice that each term of this sequence is smaller than the preceding term. This is a decreasing sequence.
Let \(( a, b )\) be any interval containing the number \(2\). Since the only condition we have placed on the interval \(( a, b )\) is that it contain \(2\), anything we conclude must be true for any interval containing \(2\), no matter how small.
We can pick \(n\) large enough to make \(\frac{1}{n}\) as small as we please, so pick \(n\) large enough that \(\frac{1}{n}\) is smaller than \(b – 2\).
Then \(2 + \frac{1}{n}\) is smaller than \(b\), so the \(n\)-th term of the sequence and all terms thereafter must lie in the interval \(( a, b )\).
So the interval \(( a, b )\) contains a tail of the sequence.
Since any interval containing \(2\) contains a tail of the sequence, \(2\) is the limit of the sequence.