91Ó°ÊÓ

Recall that a sequence \((a_n)\) converges to a limit L, written as \(\lim_{n \to \infty} a_n = L\), if for every \(\epsilon > 0\), there exists a natural number \(N\) such that for all \(n > N\), we have \(|a_n - L| < \epsilon\).

Let \((b_n)\) be the new sequence obtained by adding 50 to the first 1000 terms of the sequence \((a_n)\). We can write this as: \[ b_n = \begin{cases} a_n + 50 & \text{for} \; 1 \le n \le 1000 \\ a_n & \text{for} \; n > 1000 \end{cases} \]

Now we want to show that the limit of the new sequence, \((b_n)\), is also L. To do this, let's fix an arbitrary \(\epsilon > 0\). Since the limit of the original sequence \((a_n)\) is L, we have: There exists a natural number \(N_1\) such that for all \(n > N_1\), \(|a_n - L| < \epsilon\). Now, if we consider \(n > 1000\) (and \(n > N_1\)), then we have: \(|b_n - L| = |a_n - L|\) This is because for \(n > 1000\), the terms of the sequences \((a_n)\) and \((b_n)\) are the same. Since \(|a_n - L| < \epsilon\) for all \(n > N_1\), we also have: \(|b_n - L| < \epsilon\) So, there exists a natural number \(N_2 = max(1000, N_1)\) such that for all \(n > N_2\), \(|b_n - L| < \epsilon\). Therefore, by definition, the new sequence \((b_n)\) also converges to the same limit L, i.e., \(\lim_{n \to \infty} b_n = L\).