Limits of Sequences

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Before we begin calculus proper, we first define the limit of a sequence. Here a sequence \((a_n)\) is a function that assigns a real number to each natural number, i.e. \(a : \mathbb{N} \rightarrow \mathbb{R}\), viewed as the list of its values \(a_1, a_2, a_3, \ldots\). In §Limits of Functions we already studied how a function behaves as \(x \rightarrow \infty\), and the limit of a sequence can be thought of as the discrete version of this: the case where the variable takes only natural numbers and goes to \(n \rightarrow \infty\).

Convergence of Sequences

Definition 1 For a real sequence \((a_n)_{n=1}^\infty\) and a real number \(L\), if for every \(\epsilon > 0\) there exists a natural number \(N\) such that

\[n > N \implies \lvert a_n - L \rvert < \epsilon\]

holds, we call \(L\) the limit of \(a_n\) as \(n \rightarrow \infty\), and write \(\lim_{n\rightarrow\infty} a_n = L\).

This definition is taken almost verbatim from §Limits of Functions, ⁋Definition 14, and since a sequence has only natural numbers as its variable and goes in only one direction (\(+\infty\)), this is essentially the only way to define its limit. For example, \(a_n = 1/n \rightarrow 0\) is verified by choosing, for arbitrary \(\epsilon > 0\), an \(N > 1/\epsilon\); then for \(n > N\) we have \(1/n < 1/N < \epsilon\). A slight variant is when a sequence diverges to infinity, which is obtained by transcribing §Limits of Functions, ⁋Definition 13: for arbitrary \(M\), there exists \(N\) such that \(n > N \implies a_n > M\); for example, \(b_n = n\) is such a sequence. However, there also exist sequences that neither converge nor diverge to infinity (Example 11 (A divergent sequence)).

The basic properties of convergent sequences are mostly obtained by copying the proofs from limits of functions. For example, the following proposition can be proved in exactly the same way as §Limits of Functions, ⁋Proposition 5 (Limit Laws).

Proposition 2 (Limit Laws for Sequences) Suppose two sequences \(a_n\), \(b_n\) converge, with limits

\[\lim_{n\rightarrow\infty} a_n = L, \qquad \lim_{n\rightarrow\infty} b_n = M\]

Then

1. \(\lim (a_n + b_n) = L + M\),
2. for any constant \(c\), \(\lim c a_n = cL\),
3. \(\lim a_n b_n = LM\),
4. if \(M \neq 0\) and \(b_n \neq 0\) for all \(n\), then \(\lim a_n/b_n = L/M\)

hold.

Properties of Limits

For a sequence \((a_n)\), if there exists a positive number \(M\) such that \(\lvert a_n\rvert \leq M\) for all \(n\), we call \((a_n)\) bounded. A convergent sequence is bounded because, except for finitely many initial terms, all its terms gather near the limit point.

Proposition 3 A convergent sequence is bounded.

Proof

If \(a_n \rightarrow L\), then for \(\epsilon = 1\) there exists \(N\) such that for \(n \geq N\) we have \(\lvert a_n\rvert \leq \lvert L\rvert + 1\). Including the remaining finitely many terms, set \(M = \max\{\lvert a_1\rvert, \ldots, \lvert a_{N-1}\rvert, \lvert L\rvert + 1\}\); then \(\lvert a_n\rvert \leq M\) for all \(n\).

By the same copying, the following is the sequence version of §Limits of Functions, ⁋Proposition 8 (Squeeze Theorem).

Proposition 4 (Squeeze Theorem) If for all sufficiently large \(n\) we have \(a_n \leq c_n \leq b_n\), and \(a_n \rightarrow L\), \(b_n \rightarrow L\), then \(c_n \rightarrow L\).

Then we obtain the following simple but useful result.

Proposition 5 (Ratio Test) If a real sequence \(a_n\) satisfies \(a_n > 0\) for all \(n\), and the sequence of ratios of consecutive terms \(a_{n+1}/a_n\) converges to a value \(L\) less than \(1\), then \(a_n \rightarrow 0\).

Proof

Choose \(r\) with \(L < r < 1\); then for sufficiently large \(n \geq N\) we have \(a_{n+1}/a_n < r\), so \(a_{N+k} < r^k a_N\). Since \(0 < r < 1\), we have \(r^k \rightarrow 0\), and by the squeeze theorem \(a_n \rightarrow 0\).

Similarly, we collect in the following example several results that are frequently used in actual calculations.

Example 6 The following are basic examples of limits of sequences.

1. For \(p > 0\), \(1/n^p \rightarrow 0\) holds. Since \(n^p \geq n\) for \(n \geq 1\), we have \(0 < 1/n^p \leq 1/n \rightarrow 0\), and thus Proposition 4 (Squeeze Theorem) applies.

2. More generally, the ratio of two polynomials of the same degree is determined by the ratio of their leading coefficients.

   \[\frac{a_k n^k + \cdots}{b_k n^k + \cdots}\]

   Dividing numerator and denominator by \(n^k\), both consist of finitely many terms of the form \(1/n^j\) and a constant term. Then since \(1/n^j \rightarrow 0\), the numerator and denominator converge to their respective leading coefficients. If the denominator has higher degree than the numerator, then by Proposition 4 (Squeeze Theorem) and the previous item 1, this ratio converges to \(0\); similarly, if the numerator has higher degree than the denominator, this ratio diverges.

3. If \(\lvert r\rvert < 1\), then \(r^n \rightarrow 0\). To verify this, set \(\lvert r\rvert = 1/(1+h)\) for suitable \(h>0\). Then by the binomial theorem \((1+h)^n \geq 1 + nh\), and therefore

   \[\lvert r\rvert^n = \frac{1}{(1+h)^n} \leq \frac{1}{1+nh} \rightarrow 0\]

   The convergence of the last term used the result of item 2 above. If \(r=1\), this sequence is identically \(1\), so it trivially converges to \(1\); if \(\lvert r\rvert > 1\), similarly setting \(r=1+h\) gives

   \[\lvert r\rvert^n =(1+h)^n \geq 1+nh\]

   so no matter what \(M\) we choose, by making \(n\) sufficiently large we can make \(\lvert r\rvert^n\) exceed \(M\), and thus \(\lvert r\rvert^n\) diverges.

4. \(n^{1/n} \rightarrow 1\). To verify this, set \(n^{1/n} = 1 + h_n\) (\(h_n \geq 0\)); then by the binomial theorem

   \[n = (1+h_n)^n \geq \binom{n}{2}h_n^2 = \frac{n(n-1)}{2}h_n^2\]

   so \(h_n^2 \leq 2/(n-1) \rightarrow 0\), i.e. \(h_n \rightarrow 0\).

5. For \(r > 1\), \(p > 0\), \(n^p/r^n \rightarrow 0\) holds. This follows immediately from Proposition 5 (Ratio Test), since the ratio of consecutive terms is

   \[\frac{(n+1)^p}{r^{n+1}}\cdot\frac{r^n}{n^p} = \frac{1}{r}\left(1+\frac{1}{n}\right)^p \rightarrow \frac{1}{r} < 1\]
6. Similarly, for \(r > 1\), \(p > 0\), the ratio of consecutive terms of the sequence \(r^n/n!\) is \(r/(n+1) \rightarrow 0 < 1\), so by the same reasoning it is \(0\).

Monotone Convergence Theorem

Meanwhile, to prove the following proposition we need knowledge of analysis, so it is impossible for us to prove it at present; however, since the result itself is useful, we accept it in advance.

Proposition 7 (Monotone Convergence) An increasing sequence bounded above and a decreasing sequence bounded below converge. (The limit of an increasing sequence is the supremum of its terms.)

The most useful application of this is the existence proof of the following natural constant.

Example 8 (The natural constant \(e\)) The sequence \(a_n = (1 + 1/n)^n\) is increasing and bounded above, and thus converges by Proposition 7 (Monotone Convergence).

First let us show that this sequence is increasing. Applying the arithmetic–geometric mean inequality to \(n\) copies of \(1+1/n\) and one \(1\), the arithmetic mean is

\[\frac{n(1+1/n)+1}{n+1} = \frac{n+2}{n+1} = 1 + \frac{1}{n+1}\]

and the geometric mean is \(((1+1/n)^n\cdot 1)^{1/(n+1)}=a_n^{1/(n+1)}\), so

\[a_n \leq \left(1+\frac{1}{n+1}\right)^{n+1} = a_{n+1}\]

and moreover, since the \(n+1\) numbers are not all equal in the equality condition, we obtain the strict inequality \(a_n < a_{n+1}\).

Now to show that this sequence is bounded above, observe by the binomial theorem that

\[a_n = \sum_{k=0}^{n}\binom{n}{k}\frac{1}{n^k}\]

Each term satisfies

\[\binom{n}{k}\frac{1}{n^k} = \frac{1}{k!}\cdot\frac{n(n-1)\cdots(n-k+1)}{n^k} \leq \frac{1}{k!}\]

and since \(k! \geq 2^{k-1}\) (\(k \geq 1\)),

\[a_n \leq \sum_{k=0}^{n}\frac{1}{k!} \leq 1 + \sum_{k=1}^{\infty}\frac{1}{2^{k-1}} = 3\]

We define the limit of this sequence as the natural constant \(e = 2.718\ldots\).

Subsequences

Meanwhile, when analyzing whether a sequence converges, it is useful to consider a new sequence formed by selecting only some of its terms.

Definition 9 Given a sequence \((a_n)\) and a strictly increasing sequence of natural numbers \(n_1 < n_2 < n_3 < \cdots\), the new sequence \((a_{n_k})_{k\geq 1}\) is called a subsequence of \((a_n)\).

That is, it is obtained from the sequence \(a_n\) by selecting terms while preserving the order of indices, skipping over some terms along the way. It is intuitively obvious that if the original sequence converges to some value, then any subsequence of it also converges to that same value.

Proposition 10 If a sequence \(a_n\) converges to \(L\), then every subsequence \((a_{n_k})\) also converges to \(L\).

Proof

For arbitrary \(\epsilon > 0\), pick \(N\) such that for \(n \geq N\) we have \(\lvert a_n - L\rvert < \epsilon\). Then by definition \(n_k \geq k\), and thus for \(k \geq N\) we have \(n_k \geq N\), so

\[\lvert a_{n_k} - L\rvert < \epsilon\]

That is, \(a_{n_k} \rightarrow L\).

This proposition is more useful when showing that a sequence does not converge than when showing that it does. By its contrapositive, if there exist two subsequences with different limits, then the original sequence \((a_n)\) does not converge.

Example 11 (A divergent sequence) Consider the sequence \(a_n = (-1)^n\). The even subsequence \(a_{2k} = 1 \rightarrow 1\) and the odd subsequence \(a_{2k-1} = -1 \rightarrow -1\) have different limits, so by Proposition 10, \((a_n)\) diverges.

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References

[Ste] J. Stewart, Calculus, 8th ed., Cengage Learning, 2016.
[Kim] 김홍종, 미적분학 1·2, 제3개정판, 서울대학교출판문화원, 2020.