Proving Subsequence Convergence Without Sequence Convergence

It is a common scenario in advanced mathematical analysis to work with sequences and subsequences. One interesting and often challenging task is to prove that a subsequence converges while the original sequence does not. This article aims to explore this concept in detail with examples.

Identifying the Sequence and Subsequence

To demonstrate that a subsequence converges while the original sequence does not, we need to carefully follow a few steps. Let's start by identifying the sequence and the subsequence. Consider $a_n$ to be a sequence and $a_{n_k}$ to be a subsequence of $a_n$.

Examine the Original Sequence

The first step is to show that the original sequence $a_n$ does not converge. This can be done by demonstrating that the sequence does not approach a single limit as $n to infty$, or that it oscillates or diverges. For instance, consider the sequence $a_n (-1)^n$. This sequence oscillates between -1 and 1 and hence does not converge.

Find the Limit of the Subsequence

The next step is to identify a limit, $L$, for the subsequence $a_{n_k}$. Show that the terms of the subsequence approach $L$ as $k to infty$. To formally prove that $a_{n_k}$ converges to $L$, use the epsilon-delta definition of convergence: for every $epsilon > 0$, there exists a $K$ such that for all $k geq K$, $|a_{n_k} - L| . For example, consider the subsequence $a_{2k} 1$, the subsequence of even-indexed terms. This subsequence converges to 1 as $k to infty$.

Use the Definition of Convergence

Formally, to prove convergence, use the epsilon-delta definition. For example, if we choose $epsilon > 0$, we can show that for any given $epsilon$, there exists a $K$ such that for all $k geq K$, the condition $|a_{2k} - 1| holds. Since $a_{2k} 1$ for all $k$, this condition is always satisfied for any $epsilon > 0$ and any sufficiently large $K$.

Demonstrate Divergence of the Original Sequence

To show that the original sequence does not converge, explicitly state why it does not satisfy the convergence criteria. For instance, consider the sequence $c_n cos(npsin(frac{pi}{2}npi))$. This sequence is bounded by 2 but does not have a limit. However, consider the subsequence $c_{n_k} cos(3^kpi)sin(frac{pi}{2}3^kpi) cos(3^kpi)$. This subsequence converges to -2 as $k to infty$.

Conclusion

By following these steps, we can prove that a subsequence converges even when the original sequence does not. This method is particularly useful when dealing with bounded sequences with no further constraints.

Further Exploration

This concept is often applicable in various advanced mathematical settings. For further exploration, you may want to study more about sequences and series, particularly focusing on boundedness and subsequential limits. Here are some related terms and concepts:

Bounded Sequences: Sequences whose terms are within a certain bound, such as the sequence $c_n$. Subsequential Limits: Limits of subsequences of a given sequence. Epsilon-Delta Definition: A rigorous definition of convergence that formulates the concept of limits.

By understanding these concepts, you can better handle complex scenarios in mathematical analysis, particularly those involving bounded sequences without additional constraints.