1. **Problem statement:** Given a sequence $\{a_n\}$ of real numbers, the subsequences $\{a_{2n}\}_{n=1}^\infty$ and $\{a_{2n-1}\}_{n=1}^\infty$ both converge to the same limit $L$. We need to show that the original sequence $\{a_n\}$ also converges to $L$. 2. **Recall the definition of convergence:** A sequence $\{a_n\}$ converges to $L$ if for every $\epsilon > 0$, there exists an $N$ such that for all $n > N$, $|a_n - L| < \epsilon$. 3. **Given:** - $\lim_{n \to \infty} a_{2n} = L$ means for every $\epsilon > 0$, there exists $N_1$ such that if $n > N_1$, then $|a_{2n} - L| < \epsilon$. - $\lim_{n \to \infty} a_{2n-1} = L$ means for every $\epsilon > 0$, there exists $N_2$ such that if $n > N_2$, then $|a_{2n-1} - L| < \epsilon$. 4. **To prove:** $\lim_{n \to \infty} a_n = L$. 5. **Proof:** Let $\epsilon > 0$ be arbitrary. Since $\lim_{n \to \infty} a_{2n} = L$, choose $N_1$ such that for all $n > N_1$, $|a_{2n} - L| < \epsilon$. Since $\lim_{n \to \infty} a_{2n-1} = L$, choose $N_2$ such that for all $n > N_2$, $|a_{2n-1} - L| < \epsilon$. Let $N = \max(N_1, N_2)$. For any $n > 2N$, consider two cases: - If $n$ is even, say $n = 2k$, then $k > N$ and so $|a_n - L| = |a_{2k} - L| < \epsilon$. - If $n$ is odd, say $n = 2k - 1$, then $k > N$ and so $|a_n - L| = |a_{2k-1} - L| < \epsilon$. 6. **Conclusion:** For all $n > 2N$, $|a_n - L| < \epsilon$, which means $\lim_{n \to \infty} a_n = L$. Thus, the original sequence $\{a_n\}$ converges to the same limit $L$ as its subsequences. **Final answer:** $\boxed{\lim_{n \to \infty} a_n = L}$