[problem:Monotone Convergence And Euler's Number | 1] Define $a_n = (1 + 1/n)^n$. Prove that $(a_n)$ is increasing and bounded above by $3$, and deduce that $e = \lim_{n \to \infty} (1 + 1/n)^n$ exists. [/problem]

---

[problem:Cauchy Criterion For Series | 2] Use the [General Principle of Convergence](/theorems/172) to prove that if $\sum_{n=1}^\infty a_n$ converges, then $a_n \to 0$. Show by example that the converse fails. [/problem]

[solution] **Step 1: Convergence implies $a_n \to 0$.** If $\sum a_n$ converges, the partial sums $S_N = \sum_{n=1}^N a_n$ converge, hence are Cauchy. For any $\varepsilon > 0$, there exists $N$ with $|S_m - S_n| < \varepsilon$ for all $m > n \geq N$. Taking $m = n + 1$: $|a_{n+1}| = |S_{n+1} - S_n| < \varepsilon$ for $n \geq N$. So $a_n \to 0$. **Step 2: The converse fails.** The harmonic series $\sum 1/n$ has $a_n = 1/n \to 0$, yet diverges. The partial sums satisfy $S_{2n} - S_n = \sum_{k=n+1}^{2n} 1/k \geq n \cdot 1/(2n) = 1/2$ for all $n$, violating the Cauchy criterion with $\varepsilon = 1/2$. $\checkmark$ [/solution]

---

[problem:IVT Application — Existence Of $n$th Roots | 2] Use the [Intermediate Value Theorem](/theorems/180) to prove that for every $a > 0$ and $n \in \mathbb{N}$, there exists a unique $b > 0$ with $b^n = a$. [/problem]

[solution] **Step 1: Existence via IVT.** Define $f(x) = x^n$ on $[0, \infty)$. Since $f$ is continuous, $f(0) = 0 < a$, and $f(x) \to \infty$ as $x \to \infty$ (so there exists $M$ with $f(M) > a$), the IVT on $[0, M]$ gives $b \in (0, M)$ with $f(b) = a$. **Step 2: Uniqueness.** If $0 < b_1 < b_2$, then $b_1^n < b_2^n$ (since $x^n$ is strictly increasing on $(0, \infty)$). So there is at most one positive $n$th root. $\checkmark$ [/solution]

---

[problem:Mean Value Theorem — Derivative Zero Implies Constant | 2] Let $f: [a, b] \to \mathbb{R}$ be continuous on $[a, b]$ and differentiable on $(a, b)$ with $f'(x) = 0$ for all $x \in (a, b)$. Prove that $f$ is constant. [/problem]

[solution] Fix any $x \in (a, b]$. By the [Mean Value Theorem](/theorems/186) applied to $[a, x]$: there exists $c \in (a, x)$ with \begin{align*} f(x) - f(a) = f'(c)(x - a) = 0 \cdot (x - a) = 0. \end{align*} So $f(x) = f(a)$ for all $x \in [a, b]$, proving $f$ is constant. This simple application of the MVT has a profound consequence: antiderivatives are unique up to additive constants. If $F'(x) = G'(x)$ on $(a, b)$, then $(F - G)'(x) = 0$, so $F - G$ is constant. $\checkmark$ [/solution]

---

[problem:Riemann Integrability Of The Thomae Function | 3] Prove that the Thomae function $f: [0, 1] \to \mathbb{R}$ (defined by $f(p/q) = 1/q$ for $p/q$ in lowest terms, $f(x) = 0$ for $x$ irrational) is Riemann integrable with $\int_0^1 f = 0$. [/problem]

[solution] **Step 1: Lower integral.** Since $f \geq 0$ and $f = 0$ on the irrationals (which are dense), every subinterval $[x_{j-1}, x_j]$ has $m_j = \inf f = 0$. So $s(f, \mathcal{D}) = 0$ for all $\mathcal{D}$, giving $I_*(f) = 0$. **Step 2: Upper integral.** Fix $\varepsilon > 0$. Choose $N$ with $1/N < \varepsilon/2$. The set $B = \{x \in [0,1] : f(x) \geq 1/N\}$ consists of rationals $p/q$ with $q \leq N$, so $|B| \leq N^2$. Construct $\mathcal{D}$ placing each point of $B$ inside an interval of total length at most $\varepsilon/(2)$. On these "bad" intervals ($M_j \leq 1$), the contribution to $S(f, \mathcal{D})$ is at most $1 \cdot \varepsilon/2 = \varepsilon/2$. On all other intervals, $M_j < 1/N < \varepsilon/2$, contributing at most $(\varepsilon/2) \cdot 1 = \varepsilon/2$. So $S(f, \mathcal{D}) - s(f, \mathcal{D}) \leq \varepsilon$. Since $\varepsilon$ was arbitrary, $I^*(f) = I_*(f) = 0$. $\checkmark$ [/solution]

---

[problem:Taylor's Theorem — Smooth But Not Analytic | 3] Define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = e^{-1/x^2}$ for $x \neq 0$ and $f(0) = 0$. Prove that $f^{(k)}(0) = 0$ for all $k \geq 0$, so the Taylor series of $f$ at $0$ is identically zero, yet $f(x) > 0$ for $x \neq 0$. [/problem]

[solution] **Step 1: $f$ is smooth away from $0$.** For $x \neq 0$, $f(x) = e^{-1/x^2}$ is a composition of smooth functions, hence smooth. **Step 2: $f^{(k)}(0) = 0$ for all $k$.** We prove by induction that $f^{(k)}(0) = 0$ and that for $x \neq 0$, $f^{(k)}(x) = P_k(1/x) e^{-1/x^2}$ for some polynomial $P_k$. *Base case ($k = 0$):* $f^{(0)}(0) = f(0) = 0$ and $f^{(0)}(x) = e^{-1/x^2}$ for $x \neq 0$. *Inductive step:* Assume $f^{(k)}(0) = 0$ and $f^{(k)}(x) = P_k(1/x) e^{-1/x^2}$. Then: \begin{align*} f^{(k+1)}(0) = \lim_{h \to 0} \frac{f^{(k)}(h) - f^{(k)}(0)}{h} = \lim_{h \to 0} \frac{P_k(1/h) e^{-1/h^2}}{h}. \end{align*} For any polynomial $Q$, $\lim_{h \to 0} Q(1/h) e^{-1/h^2} = 0$ because $e^{-1/h^2}$ decays faster than any power of $|h|$ as $h \to 0$ (equivalently, $t^m e^{-t} \to 0$ as $t \to \infty$ for all $m$). So $f^{(k+1)}(0) = 0$. **Step 3: Conclude.** The Taylor series at $0$ is $\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} x^k = 0$ for all $x$. But $f(x) = e^{-1/x^2} > 0$ for all $x \neq 0$. So $f$ is $C^\infty$ but not analytic at $0$: its Taylor series converges everywhere but does not converge to $f$. The Lagrange remainder $R_n(x) = f(x) \neq 0$ for $x \neq 0$, despite $R_n(x) \to 0$ as $x \to 0$ for each fixed $n$. $\checkmark$ [/solution]