Advanced Analysis

Sequences Continued

Meta (Uniqueness of Limits)
A key idea that makes limits sensible as a thing to study is that they are unique: if you can establish that \(\lim_{n \rightarrow \infty} a_n = L\) and I can prove that \(\lim_{n \rightarrow \infty} a_n = M\), then we must agree \(L = M\). To see this, suppose \(L \neq M\) and set \(\epsilon := |L-M|/2\), which will be strictly positive. You can prove that \(|a_n - L| < \epsilon/2\) for all \(n \geq N_1\). I can prove that \(|a_n - M| < \epsilon/2\) for all \(n \geq N_2\). Choosing any \(n \geq \max\{N_1,N_2\}\) gives that we must both be correct and consequently
\[{}|L - M| \leq |(L - a_n) - (a_n - M)|{}\]
\[{}\leq |a_n - L| + |a_n - M|{}\]
\[{}< \frac{\epsilon}{2} + \frac{\epsilon}{2}{}\]
\[{}= \epsilon = \frac{1}{2} |L-M|. {}\]
If \(|L-M|\) is strictly positive, this gives a contradiction, since it would imply that \(|L-M|/2 < 0\) and hence \(|L-M| < 0\). Thus \(|L-M| = 0\), meaning \(L=M\).
Example (Proving limits exist)
Suppose that \(\{x_n\}_{n=1}^\infty\) converges to \(L\) as \(n \rightarrow \infty\). For each \(n\), let \(y_n := \frac{1}{n}(x_1 + \cdots + x_n)\). Then \(\{y_n\}_{n=1}^\infty\) is also convergent to \(L\) as \(n \rightarrow \infty\).
Proof
Meta (Main Ideas)
To compare \(y_n\) to \(L\), we need to explicitly reduce to expressions involving \(|x_n - L|\) because that's the only thing we know must be small. One approach looks like this:
\[{}\left| \frac{x_1+\cdots+x_n}{n} - L \right|{}\]
\[{}= \left| \frac{x_1+\cdots+x_n}{n} - \frac{nL}{n} \right|{}\]
\[{}= \left| \frac{(x_1-L)+\cdots+(x_n-L)}{n} \right|{}\]
\[{}\leq \frac{1}{n} (|x_1-L| + \cdots + |x_n - L|){}\]
The good news: For large \(n\), the terms \(|x_n - L|\) are very small.

The bad news: Not all of the indices in our expression are large.

The good news: As we let \(n\) get extremely large, the proportion of terms that are small grows to be almost all of them, and that's enough for us to proceed.

The steps must be carefully arranged in the proper order:
  1. Fix an \(\epsilon > 0\).
  2. We know that there is some \(N_0\) such that \(|x_n - L| < \epsilon/2\) when \(n \geq N_0\).
  3. We know that convergent sequences are bounded. This means that there is some constant \(C\) such that \(|x_n - L| \leq C\) for all \(n\), not just \(n \geq N_0\). Without loss of generality, \(C > 0\).
  4. Take some value of \(n\) much larger than \(N_0\) to be determined:
    \[{}\frac{1}{n} \sum_{j=1}^n |x_j - L|{}\]
    \[{}= \frac{1}{n} \sum_{j=1}^{N_0-1} |x_j - L| {}\]
    \[{}+ \frac{1}{n} \sum_{j=N_0}^n |x_j - L|{}\]
    \[{}< \frac{1}{n} \sum_{j=1}^{N_0-1} C + \frac{1}{n} \sum_{j=N_0}^n \frac{\epsilon}{2}{}\]
    \[{}\leq \frac{(N_0-1)C}{n} + \frac{n-N_0}{n} \frac{\epsilon}{2}{}\]
    \[{}\leq \frac{(N_0-1)C}{n} + \frac{\epsilon}{2}.{}\]
  5. Constrain
    \[n \geq \max\{N_0, 2(N_0-1)C \epsilon^{-1}\}.\]
    All \(n\) in this range are big enough for the derivation above to be valid. They're also big enough that the first term is no greater than \(\epsilon/2\).

1. Limits and Inequalities

Theorem (Inequalities)
Suppose that \(\{a_n\}_{n=1}^\infty\) is a convergent sequence.
  • If \(a_n \geq M\) for all \(n\) sufficiently large, then
    \[ \lim_{n \rightarrow \infty} a_n \geq M. \]
  • If \(a_n \leq M\) for all \(n\) sufficiently large, then
    \[ \lim_{n \rightarrow \infty} a_n \leq M. \]
Proof
Meta (Main Idea)
For the first inequality, suppose that \(L\) is the limit of the \(a_n\)'s and that \(L < M\). This means we can take \(\epsilon := M - L\) because it's positive. There is some \(N_0\) such that \(a_n \geq M\) for all \(n \geq N_0\), and there is some \(N\) such that \(|a_n - L| < \epsilon\) for all \(n \geq N\). Choose any \(n \geq \max\{N_0,N\}\). We have that \(a_n \geq M\) and \(-\epsilon < a_n - L < \epsilon\), the latter implying that \(L - \epsilon < a_n < L + \epsilon\) by adding \(L\) to each part. But \(L + \epsilon = M\), so we have that \(a_n \geq M\) and \(a_n < M\), contradicting the order trichotomy. Thus the assumed inequality \(L < M\) must be false.
Meta (Limits and Strict Inequalities)
Limits do not respect strict inequalities. For example, the sequence \(\{\frac{1}{n}\}_{n=1}^\infty\) is a sequence of strictly positive terms, but the limit is not strictly positive (but rather is \(0\)). So the most that can be said is that if the terms satisfy a strict inequality, the limit will satisfy the non-strict counterpart.

2. Diverging to Infinity

Definition
We say that a sequence \(\{x_n\}_{n=1}^\infty\) diverges to \(+\infty\) when for each \(M\), there exists \(N \in {\mathbb N}\) such that \(x_n > M\) for all \(n \geq N\). For example, the sequence \(\{n^2\}_{n=1}^\infty\) diverges to \(+\infty\). Note that this is just one specific kind of divergent sequence among many; a general divergent sequence is one which bounces around in no discernible way and so we don't study them much because of the extreme lack of structure.