1. **Problem statement:** We have a sequence $\{a_n\}$ of real numbers such that the subsequences $\{a_{2n}\}$, $\{a_{2n-1}\}$, and $\{a_{3n}\}$ all converge. We need to show that the entire sequence $\{a_n\}$ converges. 2. **Recall the definition of convergence:** A sequence $\{a_n\}$ converges to a limit $L$ if for every $\epsilon > 0$, there exists $N$ such that for all $n > N$, $|a_n - L| < \epsilon$. 3. **Given:** - $\{a_{2n}\} \to L_2$ - $\{a_{2n-1}\} \to L_1$ - $\{a_{3n}\} \to L_3$ 4. **Goal:** Show $L_1 = L_2 = L_3 = L$ and that $\{a_n\} \to L$. 5. **Step 1: Use the subsequence $\{a_{3n}\}$ which is a subsequence of both $\{a_{2n}\}$ and $\{a_{2n-1}\}$ subsequences.** - Note that $\{a_{3n}\}$ is a subsequence of $\{a_n\}$ indexed by multiples of 3. - Among these multiples of 3, some are even and some are odd: - If $3n$ is even, then $a_{3n}$ is in $\{a_{2n}\}$. - If $3n$ is odd, then $a_{3n}$ is in $\{a_{2n-1}\}$. 6. **Step 2: Check parity of $3n$:** - Since 3 is odd, $3n$ is even if and only if $n$ is even. - $3n$ is odd if and only if $n$ is odd. 7. **Step 3: Consider subsequences of $\{a_{3n}\}$:** - $\{a_{6k}\} = \{a_{2(3k)}\}$ is a subsequence of $\{a_{2n}\}$ and converges to $L_2$. - $\{a_{3(2k-1)}\} = \{a_{6k-3}\} = \{a_{2(3k-2)+1}\}$ is a subsequence of $\{a_{2n-1}\}$ and converges to $L_1$. 8. **Step 4: Since $\{a_{3n}\}$ converges to $L_3$, both subsequences $\{a_{6k}\}$ and $\{a_{6k-3}\}$ must converge to $L_3$ as well (because they are subsequences of $\{a_{3n}\}$).** 9. **Step 5: From step 7 and 8, we have:** - $\{a_{6k}\} \to L_2$ and $\{a_{6k}\} \to L_3$ so $L_2 = L_3$. - $\{a_{6k-3}\} \to L_1$ and $\{a_{6k-3}\} \to L_3$ so $L_1 = L_3$. 10. **Step 6: Therefore, $L_1 = L_2 = L_3 = L$ for some limit $L$. 11. **Step 7: Since $\{a_{2n}\}$ and $\{a_{2n-1}\}$ both converge to $L$, the entire sequence $\{a_n\}$ converges to $L$ because every term of $\{a_n\}$ is either in $\{a_{2n}\}$ or $\{a_{2n-1}\}$. **Final answer:** The sequence $\{a_n\}$ converges to the common limit $L$ of its subsequences $\{a_{2n}\}$, $\{a_{2n-1}\}$, and $\{a_{3n}\}$.