September 29th, 2026

Updated 1 Oct 2026Notes (PDF)

[0.0.15]Definition(Subsequence)#

Let $(x_n)_{n = 1}^\infty$ be a sequence. A subsequence of $(x_n)$ is a sequence $(x_{n_k})_{k = 1}^\infty$ such that $(n_k)_{k = 1}^\infty$ is a strictly increasing sequence of positive integers.

[0.0.16]Lemma#

If $(x_{n_k})$ is a subsequence of $(x_n)$ , then $n_k \ge k$ for every $k \in \bZ_+$ . In particular, $n_k \to \infty$ .

Proof.

We proceed by induction. Since $n_1 \in \bZ_+$ , we have $n_1 \ge 1$ . Suppose $n_k \ge k$ . Since the indices are strictly increasing, $n_{k + 1} > n_k$ , and thus $n_{k + 1} \ge n_k + 1 \ge k + 1$ . Therefore $n_k \ge k$ for every $k \in \bZ_+$ .

It follows immediately that, for every $N \in \bZ_+$ , we have $n_k \ge N$ whenever $k \ge N$ . Thus $n_k \to \infty$ .

[0.0.17]Theorem(Subsequences Preserve Limits)#

Suppose $x_n \to x$ . Then every subsequence $(x_{n_k})$ also converges to $x$ .

Proof.

Fix $\epsilon > 0$ . Since $x_n \to x$ , there is some $N \in \bZ_+$ such that $|x_n - x| < \epsilon$ for every $n \ge N$ . Since $n_k \ge k$ , we have $n_k \ge N$ whenever $k \ge N$ . Thus $|x_{n_k} - x| < \epsilon$ for every $k \ge N$ , and so $x_{n_k} \to x$ .

The contrapositive of the previous theorem is often useful: if some subsequence of $(x_n)$ does not converge to $x$ , then $(x_n)$ itself does not converge to $x$ .

There is also a useful way to construct a subsequence witnessing failure of convergence.

[0.0.18]Theorem#

Suppose $x_n$ does not converge to $x$ . Then there is some $\epsilon > 0$ and some subsequence $(x_{n_k})$ such that $|x_{n_k} - x| \ge \epsilon$ for every $k \in \bZ_+$ .

Proof.

Since $x_n$ does not converge to $x$ , there is some $\epsilon > 0$ such that, for every $N \in \bZ_+$ , there exists some $n \ge N$ for which $|x_n - x| \ge \epsilon$ .

Choose $n_1$ such that $|x_{n_1} - x| \ge \epsilon$ . Having chosen $n_k$ , choose $n_{k + 1} > n_k$ such that $|x_{n_{k + 1}} - x| \ge \epsilon$ . This produces a subsequence $(x_{n_k})$ satisfying $|x_{n_k} - x| \ge \epsilon$ for every $k$ .

[0.0.19]Definition(Subsequential Limit)#

Let $(x_n)$ be a sequence of reals. We say $x \in \bR$ is a subsequential limit of $(x_n)$ if there is some subsequence $(x_{n_k})$ such that $x_{n_k} \to x$ .

Thus, if $x_n \to x$ , then $x$ is the only possible subsequential limit of $(x_n)$ .

[0.0.20]Theorem(Bolzano–Weierstrass)#

Every bounded sequence of reals has a convergent subsequence.

Proof.

Suppose $(x_n)$ is bounded. Then there is some $M > 0$ such that $x_n \in [-M, M]$ for every $n \in \bZ_+$ . Let $I_0 = [-M, M]$ .

Bisect $I_0$ into two closed intervals. At least one of these intervals contains infinitely many terms of $(x_n)$ ; otherwise, $I_0$ itself would contain only finitely many terms. Let $I_1$ be one such interval.

Continuing inductively, suppose $I_k$ has been chosen and contains infinitely many terms of $(x_n)$ . Bisect $I_k$ into two closed intervals, and let $I_{k + 1}$ be one of the halves containing infinitely many terms. We therefore obtain a nested sequence of closed intervals

I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots

such that each $I_k$ contains infinitely many terms of $(x_n)$ and the length of $I_k$ is $2M/2^k$ .

Write $I_k = [a_k, b_k]$ . Since the intervals are nested, $(a_k)$ is monotonically increasing and $(b_k)$ is monotonically decreasing. Both are bounded, and so there exist $a, b \in \bR$ such that $a_k \to a$ and $b_k \to b$ . Moreover,

0 \le b_k - a_k = \frac{2M}{2^k} \to 0,

and thus $a = b$ . Let $x = a = b$ .

Since each $I_k$ contains infinitely many terms of $(x_n)$ , we may choose $n_1 < n_2 < \cdots$ such that $x_{n_k} \in I_k$ for every $k$ . Then $a_k \le x_{n_k} \le b_k$ , and since both $a_k$ and $b_k$ converge to $x$ , the Squeeze Theorem gives $x_{n_k} \to x$ .

[0.0.21]Corollary#

Every bounded sequence of reals has at least one subsequential limit.

We now introduce a way of describing the asymptotic upper and lower behavior of a bounded sequence. Let $x_n$ be any bounded sequence, and define $S_k = \{x_n : n \ge k\}$ to be the tail of the sequence after removing its first $k - 1$ terms. Since $x_n$ is bounded, each $S_k$ is nonempty and bounded, and so we may define $U_k = \sup(S_k)$ and $L_k = \inf(S_k)$ .

Since $S_{k + 1} \subseteq S_k$ , the sequence $(U_k)$ is monotonically decreasing and the sequence $(L_k)$ is monotonically increasing. Both are bounded, and so both converge. Intuitively, $U_k$ records the largest values that remain possible arbitrarily far into the sequence, while $L_k$ records the smallest values that remain possible arbitrarily far into the sequence.

[0.0.22]Definition(Limit Superior and Limit Inferior)#

Suppose $(x_n)$ is a bounded sequence of reals. We define its limit superior and limit inferior by

\limsup_{n \to \infty} x_n = \lim_{k \to \infty} \sup_{n \ge k} x_n = \inf_{k \ge 1} \sup_{n \ge k} x_n,

and

\liminf_{n \to \infty} x_n = \lim_{k \to \infty} \inf_{n \ge k} x_n = \sup_{k \ge 1} \inf_{n \ge k} x_n.

For intuition, consider $x_n = \sin(\pi n/2)$ . Then

x_n = 1, 0, -1, 0, 1, 0, -1, 0, \dots.

Every tail contains both $1$ and $-1$ , and so $\sup_{n \ge k} x_n = 1$ and $\inf_{n \ge k} x_n = -1$ for every $k$ . Thus $\limsup(x_n) = 1$ and $\liminf(x_n) = -1$ . These values are realized by subsequences: $x_{4k + 1} \to 1$ and $x_{4k + 3} \to -1$ . There is also the subsequential limit $0$ , obtained from $x_{2k}$ . Thus $\limsup$ and $\liminf$ should be thought of as the largest and smallest subsequential limits.

[0.0.23]Theorem(Subsequential Characterization of

\limsup

and

\liminf

Suppose $(x_n)$ is a bounded sequence of reals. Then $\limsup(x_n)$ is the largest subsequential limit of $(x_n)$ , and $\liminf(x_n)$ is the smallest subsequential limit of $(x_n)$ .

Proof.

Let $U_k = \sup_{n \ge k} x_n$ and $L_k = \inf_{n \ge k} x_n$ , and write $U = \limsup(x_n)$ and $L = \liminf(x_n)$ .

We first show that $U$ is a subsequential limit. Choose $n_1$ such that $x_{n_1} > U_1 - 1$ . Having chosen $n_{k - 1}$ , let $m_k = n_{k - 1} + 1$ . Since $U_{m_k} = \sup_{n \ge m_k} x_n$ , there exists some $n_k \ge m_k$ such that

U_{m_k} - \frac{1}{k} < x_{n_k} \le U_{m_k}.

Then $n_k > n_{k - 1}$ , so $(x_{n_k})$ is a subsequence. Since $m_k \to \infty$ , we have $U_{m_k} \to U$ , and therefore the Squeeze Theorem gives $x_{n_k} \to U$ . Thus $U$ is a subsequential limit.

Similarly, we may choose a subsequence $(x_{r_k})$ such that

L_{s_k} \le x_{r_k} < L_{s_k} + \frac{1}{k}

for some sequence $s_k \to \infty$ . Since $L_{s_k} \to L$ , it follows that $x_{r_k} \to L$ . Thus $L$ is also a subsequential limit.

It remains to show that these are the largest and smallest such limits. Suppose $x_{n_j} \to y$ for some subsequence. Fix $k \in \bZ_+$ . Since $n_j \to \infty$ , there is some $J$ such that $n_j \ge k$ for every $j \ge J$ . Thus

L_k \le x_{n_j} \le U_k

for every $j \ge J$ . Taking $j \to \infty$ gives $L_k \le y \le U_k$ . Finally, taking $k \to \infty$ gives $L \le y \le U$ . Thus every subsequential limit lies between $\liminf(x_n)$ and $\limsup(x_n)$ .

[0.0.24]Theorem(Characterization of Convergence by

\limsup

and

\liminf

Suppose $(x_n)$ is a bounded sequence of reals. Then $x_n$ converges if and only if $\limsup(x_n) = \liminf(x_n)$ . In this case,

\lim_{n \to \infty} x_n = \limsup_{n \to \infty} x_n = \liminf_{n \to \infty} x_n.

Proof.

$(\Longrightarrow)$ : Suppose $x_n \to x$ . By the previous theorem, $\limsup(x_n)$ and $\liminf(x_n)$ are both subsequential limits of $(x_n)$ . But every subsequence of a convergent sequence converges to the same limit $x$ . Thus $\limsup(x_n) = \liminf(x_n) = x$ .

$(\Longleftarrow)$ : Suppose $\limsup(x_n) = \liminf(x_n) = x$ . Let $U_n = \sup_{m \ge n} x_m$ and $L_n = \inf_{m \ge n} x_m$ . Then $L_n \le x_n \le U_n$ for every $n$ . Since $L_n \to x$ and $U_n \to x$ , the Squeeze Theorem gives $x_n \to x$ .

[0.0.25]Definition(Cauchy Sequence)#

Let $(x_n)_{n = 1}^\infty$ be a sequence of reals. We say $(x_n)$ is Cauchy if, for every $\epsilon > 0$ , there exists some $N \in \bZ_+$ such that $|x_n - x_m| < \epsilon$ for every $n, m \ge N$ .

The Cauchy condition says that the terms of the sequence eventually become arbitrarily close to each other, without requiring that we already know what the limit is.

[0.0.26]Theorem#

Every convergent sequence of reals is Cauchy.

Proof.

Suppose $x_n \to x$ . Fix $\epsilon > 0$ . There exists some $N \in \bZ_+$ such that $|x_n - x| < \epsilon/2$ for every $n \ge N$ . Thus, for any $n, m \ge N$ ,

|x_n - x_m| \le |x_n - x| + |x_m - x| < \epsilon/2 + \epsilon/2 = \epsilon.

Therefore $(x_n)$ is Cauchy.

[0.0.27]Lemma#

Every Cauchy sequence of reals is bounded.

Proof.

Suppose $(x_n)$ is Cauchy. Taking $\epsilon = 1$ , there exists some $N \in \bZ_+$ such that $|x_n - x_m| < 1$ for every $n, m \ge N$ . In particular, taking $m = N$ , we have $|x_n - x_N| < 1$ for every $n \ge N$ . Therefore

|x_n| \le |x_n - x_N| + |x_N| < 1 + |x_N|

for every $n \ge N$ .

We may therefore choose

M = \max\{|x_1|, \dots, |x_{N - 1}|, |x_N| + 1\},

and then $|x_n| \le M$ for every $n \in \bZ_+$ . Thus $(x_n)$ is bounded.

[0.0.28]Theorem(Cauchy Completeness of

\bR

Every Cauchy sequence of reals converges.

Proof.

Suppose $(x_n)$ is Cauchy. By the previous lemma, $(x_n)$ is bounded. Thus, by the Bolzano–Weierstrass Theorem, there exists some subsequence $(x_{n_k})$ converging to some $x \in \bR$ .

Fix $\epsilon > 0$ . Since $(x_n)$ is Cauchy, there exists some $N \in \bZ_+$ such that $|x_n - x_m| < \epsilon/2$ for every $n, m \ge N$ . Since $x_{n_k} \to x$ , there exists some $K \in \bZ_+$ such that $|x_{n_k} - x| < \epsilon/2$ for every $k \ge K$ .

Choose some $k \ge \max\{K, N\}$ . Then $n_k \ge k \ge N$ . Hence, for every $n \ge N$ ,

|x_n - x| \le |x_n - x_{n_k}| + |x_{n_k} - x| < \epsilon/2 + \epsilon/2 = \epsilon.

Thus $x_n \to x$ .

The same result may be proved directly from the limit superior and limit inferior, without using Bolzano–Weierstrass.

Proof.

Suppose $(x_n)$ is Cauchy. Since every Cauchy sequence is bounded, $\limsup(x_n)$ and $\liminf(x_n)$ are defined. Let

U_k = \sup_{n \ge k} x_n \quad \text{and} \quad L_k = \inf_{n \ge k} x_n.

Fix $\epsilon > 0$ . Since $(x_n)$ is Cauchy, there exists some $N \in \bZ_+$ such that $|x_n - x_m| < \epsilon$ for every $n, m \ge N$ . Thus $x_n < x_m + \epsilon$ for every $n, m \ge N$ . Fixing $m \ge N$ and taking the supremum over $n \ge N$ gives $U_N \le x_m + \epsilon$ . Taking the infimum over $m \ge N$ then gives

U_N \le L_N + \epsilon.

Hence $0 \le U_N - L_N \le \epsilon$ .

The same argument applies to every $k \ge N$ , and therefore $U_k - L_k \to 0$ . Since $U_k \to \limsup(x_n)$ and $L_k \to \liminf(x_n)$ , we obtain

\limsup_{n \to \infty} x_n - \liminf_{n \to \infty} x_n = 0.

Thus $\limsup(x_n) = \liminf(x_n)$ , and so $(x_n)$ converges.

Combining the two directions gives the following characterization.

[0.0.29]Corollary#

A sequence of reals converges if and only if it is Cauchy.

[0.0.30]Definition(Infinite Series)#

Let $(a_n)_{n = 1}^\infty$ be a sequence of reals. For each $N \in \bZ_+$ , define the $N$ -th partial sum by $S_N = \sum_{n = 1}^N a_n$ . We say the infinite series $\sum_{n = 1}^\infty a_n$ converges to $S$ if $S_N \to S$ . If the sequence of partial sums does not converge, we say the series diverges.

[0.0.31]Theorem(Cauchy Criterion for Series)#

The series $\sum_{n = 1}^\infty a_n$ converges if and only if, for every $\epsilon > 0$ , there exists some $N \in \bZ_+$ such that

\left|\sum_{j = m + 1}^n a_j\right| < \epsilon

for every $n > m \ge N$ .

Proof.

Let $S_N = \sum_{n = 1}^N a_n$ be the sequence of partial sums. The series converges if and only if $(S_N)$ converges, which by Cauchy completeness is equivalent to $(S_N)$ being Cauchy. But, for $n > m$ ,

S_n - S_m = \sum_{j = m + 1}^n a_j.

Thus the Cauchy condition for $(S_N)$ is exactly the stated condition.

[0.0.32]Lemma#

If $|x| < 1$ , then $x^n \to 0$ .

Proof.

If $x = 0$ , the result is immediate. Suppose instead $0 < |x| < 1$ , and let $a = |x|^{-1} - 1 > 0$ . Then $|x|^{-1} = 1 + a$ . By the binomial theorem,

(1 + a)^n \ge 1 + na.

Therefore

0 < |x|^n = \frac{1}{(1 + a)^n} \le \frac{1}{1 + na}.

Since $1/(1 + na) \to 0$ , the Squeeze Theorem gives $|x|^n \to 0$ , and hence $x^n \to 0$ .

[0.0.33]Lemma(Finite Geometric Sum)#

If $x \ne 1$ , then $\sum_{n = 0}^N x^n = \frac{1 - x^{N + 1}}{1 - x}$ .

Proof.

Let $S_N = 1 + x + x^2 + \cdots + x^N$ . Then

xS_N = x + x^2 + \cdots + x^N + x^{N + 1}.

Subtracting gives $(1 - x)S_N = 1 - x^{N + 1}$ . Since $x \ne 1$ , division by $1 - x$ gives $S_N = \frac{1 - x^{N + 1}}{1 - x}$ .

[0.0.34]Theorem(Geometric Series)#

If $|x| < 1$ , then

\sum_{n = 0}^\infty x^n = \frac{1}{1 - x}.

Proof.

Let $S_N = \sum_{n = 0}^N x^n$ . By the finite geometric sum identity,

S_N = \frac{1 - x^{N + 1}}{1 - x}.

Since $|x| < 1$ , we have $x^{N + 1} \to 0$ . Therefore $S_N \to \frac{1}{1 - x}$ , and so $\sum_{n = 0}^\infty x^n = \frac{1}{1 - x}$ .

[0.0.35]Theorem(Convergence of Nonnegative Series)#

Let $(a_n)_{n = 1}^\infty$ be a sequence of nonnegative reals. Then $\sum_{n = 1}^\infty a_n$ converges if and only if its sequence of partial sums is bounded above.

Proof.

Let $S_N = \sum_{n = 1}^N a_n$ . Since $a_n \ge 0$ for every $n$ , we have $S_{N + 1} = S_N + a_{N + 1} \ge S_N$ , and so $(S_N)$ is monotonically increasing.

If $(S_N)$ is bounded above, then the Monotone Convergence Theorem implies that $(S_N)$ converges, and therefore $\sum_{n = 1}^\infty a_n$ converges.

Conversely, if $\sum_{n = 1}^\infty a_n$ converges, then $(S_N)$ converges, and every convergent sequence is bounded. Thus $(S_N)$ is bounded above.

If $a_n \ge 0$ for every $n$ and the partial sums are not bounded above, then, since they are monotonically increasing, $S_N \to +\infty$ .

[0.0.36]Definition(Divergence to

+\infty

Let $(a_n)_{n = 1}^\infty$ be a sequence of nonnegative reals. We say $\sum_{n = 1}^\infty a_n = +\infty$ if, for every $B \in \bR$ , there exists some $N \in \bZ_+$ such that $\sum_{n = 1}^N a_n > B$ .

[0.0.37]Theorem(Comparison Test)#

Suppose $0 \le a_n \le b_n$ for every $n \in \bZ_+$ . If $\sum_{n = 1}^\infty b_n$ converges, then $\sum_{n = 1}^\infty a_n$ converges.

Proof.

Let $A_N = \sum_{n = 1}^N a_n$ and $B_N = \sum_{n = 1}^N b_n$ . Since $a_n \ge 0$ , $(A_N)$ is monotonically increasing. Moreover,

A_N \le B_N \le \sum_{n = 1}^\infty b_n

for every $N$ . Thus $(A_N)$ is bounded above. By the convergence criterion for nonnegative series, $\sum_{n = 1}^\infty a_n$ converges.

[0.0.38]Definition(Absolute Convergence)#

We say the series $\sum_{n = 1}^\infty a_n$ converges absolutely if $\sum_{n = 1}^\infty |a_n|$ converges.

[0.0.39]Theorem(Absolute Convergence Implies Convergence)#

If $\sum_{n = 1}^\infty |a_n|$ converges, then $\sum_{n = 1}^\infty a_n$ converges.

Proof.

Suppose $\sum_{n = 1}^\infty |a_n|$ converges. Fix $\epsilon > 0$ . By the Cauchy criterion for series, there exists some $N \in \bZ_+$ such that

\sum_{j = m + 1}^n |a_j| < \epsilon

for every $n > m \ge N$ . By the triangle inequality,

\left|\sum_{j = m + 1}^n a_j\right| \le \sum_{j = m + 1}^n |a_j| < \epsilon.

Thus $\sum_{n = 1}^\infty a_n$ satisfies the Cauchy criterion and therefore converges.

[0.0.40]Lemma#

For every $n \in \bZ_+$ , we have $n! \ge 2^{n - 1}$ .

Proof.

We proceed by induction. For $n = 1$ , we have $1! = 1 = 2^0$ . Suppose $n! \ge 2^{n - 1}$ . Then

(n + 1)! = (n + 1)n! \ge 2n! \ge 2 \cdot 2^{n - 1} = 2^n.

Thus the result holds for every $n \in \bZ_+$ .

[0.0.41]Theorem#

The series $\sum_{n = 1}^\infty \frac{1}{n!}$ converges.

Proof.

By the previous lemma, $n! \ge 2^{n - 1}$ for every $n \in \bZ_+$ . Thus

0 \le \frac{1}{n!} \le \frac{1}{2^{n - 1}}.

But $\sum_{n = 1}^\infty \frac{1}{2^{n - 1}}$ is a convergent geometric series. Therefore, by the Comparison Test, $\sum_{n = 1}^\infty \frac{1}{n!}$ converges.

[0.0.42]Definition(Euler's Number)#

We define Euler's number $e$ by

e = \sum_{n = 0}^\infty \frac{1}{n!}.

[0.0.43]Theorem(Exponential Series)#

For every $x \in \bR$ , the series $\sum_{n = 0}^\infty \frac{x^n}{n!}$ converges absolutely.

Proof.

Fix $x \in \bR$ , and let $a_n = \frac{|x|^n}{n!}$ . If $x = 0$ , the result is immediate. Suppose $x \ne 0$ .

By the Archimedean property, choose some $N \in \bZ_+$ such that $N > 2|x|$ . Then, for every $n \ge N$ ,

a_{n + 1} = a_n \frac{|x|}{n + 1} < \frac{a_n}{2}.

It follows inductively that $a_{N + k} \le a_N/2^k$ for every $k \ge 0$ . Therefore

0 \le \sum_{k = 0}^M a_{N + k} \le a_N \sum_{k = 0}^M \frac{1}{2^k} \le 2a_N.

Thus the partial sums of the nonnegative series $\sum_{k = 0}^\infty a_{N + k}$ are bounded above, so the tail converges. Adding the finitely many terms preceding it, we obtain that $\sum_{n = 0}^\infty a_n$ converges.

Therefore $\sum_{n = 0}^\infty \frac{|x|^n}{n!}$ converges, and hence $\sum_{n = 0}^\infty \frac{x^n}{n!}$ converges absolutely.