To prove that the sequence $\{b_n\}_{n=1}^\infty$ converges to $A$ if $\{a_n\}_{n=1}^\infty$ converges to $A$, we’ll proceed by using the $\epsilon$-definition of convergence and a proof by contradiction.

Proof

1. Assume $\{a_n\}_{n=1}^\infty$ converges to $A$:
   By definition of convergence, for any $\epsilon > 0$, there exists an integer $N$ such that for all $n \geq N$, we have: $|a_n - A| < \epsilon.$

2. Define the sequence $\{b_n\}_{n=1}^\infty$:
   The sequence $\{b_n\}_{n=1}^\infty$ is defined by: $b_n = \frac{a_n + a_{n+1}}{2}.$

3. Goal:
   We want to prove that $\{b_n\}_{n=1}^\infty$ converges to $A$ as well, meaning we need to show that for any $\epsilon > 0$, there exists an integer $M$ such that for all $n \geq M$, we have: $|b_n - A| < \epsilon.$

4. Proof by Contradiction:
   Suppose, for the sake of contradiction, that $\{b_n\}_{n=1}^\infty$ does not converge to $A$. This would mean that there exists an $\epsilon > 0$ such that for every integer $M$, there exists some $n \geq M$ with: $|b_n - A| \geq \epsilon.$

5. Using the Definition of $b_n$:
   Since $b_n = \frac{a_n + a_{n+1}}{2}$, we have: $|b_n - A| = \left|\frac{a_n + a_{n+1}}{2} - A\right|.$ Expanding this, we get: $|b_n - A| = \left|\frac{(a_n - A) + (a_{n+1} - A)}{2}\right|.$ Using the triangle inequality, we can bound this as follows: $|b_n - A| \leq \frac{|a_n - A| + |a_{n+1} - A|}{2}.$

6. Applying the Convergence of $\{a_n\}_{n=1}^\infty$:
   Since $\{a_n\}_{n=1}^\infty$ converges to $A$, we know that for any $\epsilon > 0$, there exists an integer $N$ such that for all $n \geq N$, both $|a_n - A| < \epsilon$ and $|a_{n+1} - A| < \epsilon$.

   Therefore, for all $n \geq N$, we have: $|b_n - A| \leq \frac{|a_n - A| + |a_{n+1} - A|}{2} < \frac{\epsilon + \epsilon}{2} = \epsilon.$

7. Contradiction:
   This contradicts our assumption that there exists an $\epsilon > 0$ for which $|b_n - A| \geq \epsilon$ infinitely often. Hence, our assumption that $\{b_n\}_{n=1}^\infty$ does not converge to $A$ must be false.

Conclusion

Therefore, $\{b_n\}_{n=1}^\infty$ converges to $A$.

Would you like further details or clarifications? Here are five related questions for further exploration:

1. How does the sequence $\{b_n\}_{n=1}^\infty$ behave when $\{a_n\}_{n=1}^\infty$ is a constant sequence?
2. What if we defined $b_n$ as $\frac{a_n + a_{n+k}}{2}$ for some $k > 1$? Would the convergence proof hold?
3. How would this proof differ if we used a subsequence $a_{2n}$ instead of $a_{n+1}$ in defining $b_n$?
4. How might we modify this approach if $\{a_n\}$ converges to a different limit than $A$?
5. What role does the triangle inequality play in ensuring the convergence of $\{b_n\}_{n=1}^\infty$?

Tip: When handling limits in proofs, using inequalities and breaking down complex terms often simplifies proving convergence.