[problem:Computing Radius Of Convergence | 1] Find the radius of convergence of $\sum_{n=1}^{\infty} \frac{n^2}{3^n} x^n$. [/problem]

[solution] Using the ratio test: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)^2}{3^{n+1}} \cdot \frac{3^n}{n^2} = \frac{(n+1)^2}{3n^2} \to \frac{1}{3}$ as $n \to \infty$. So $1/R = 1/3$, giving $R = 3$. Alternatively, by Cauchy-Hadamard: $|a_n|^{1/n} = (n^2/3^n)^{1/n} = n^{2/n}/3 \to 1/3$ (since $n^{2/n} = e^{(2\log n)/n} \to e^0 = 1$). So $1/R = 1/3$ and $R = 3$. The series converges absolutely for $|x| < 3$ and diverges for $|x| > 3$. At $x = 3$: $\sum n^2$ diverges. At $x = -3$: $\sum (-1)^n n^2$ diverges. So the interval of convergence is $(-3, 3)$. $\checkmark$ [/solution]

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[problem:The Exponential Functional Equation | 2] Using only the power series definition $\exp(x) = \sum_{n=0}^{\infty} x^n/n!$, prove that $\exp(x + y) = \exp(x) \exp(y)$ for all $x, y \in \mathbb{R}$. [/problem]

[solution] **Step 1: The Cauchy product.** Both $\sum x^n/n!$ and $\sum y^n/n!$ converge absolutely for all $x, y$. The Cauchy product of two absolutely convergent series $\sum a_n$ and $\sum b_n$ is $\sum c_n$ where $c_n = \sum_{k=0}^n a_k b_{n-k}$, and it converges to $(\sum a_n)(\sum b_n)$. **Step 2: Compute the product.** Set $a_n = x^n/n!$ and $b_n = y^n/n!$. The $n$th Cauchy coefficient is: \begin{align*} c_n = \sum_{k=0}^{n} \frac{x^k}{k!} \cdot \frac{y^{n-k}}{(n-k)!} = \frac{1}{n!} \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k} = \frac{(x+y)^n}{n!}, \end{align*} where we used $\binom{n}{k} = n!/(k!(n-k)!)$ and the binomial theorem. **Step 3: Conclude.** $\exp(x) \exp(y) = \sum_{n=0}^{\infty} c_n = \sum_{n=0}^{\infty} \frac{(x+y)^n}{n!} = \exp(x + y)$. This derivation illustrates the power of the series approach: the functional equation follows from a single algebraic identity (the binomial theorem) applied to the Cauchy product — no limits, no continuity arguments, just formal algebra validated by absolute convergence. $\checkmark$ [/solution]

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[problem:Radius From The Complex Perspective | 2] Find the radius of convergence of the Taylor series of $f(x) = 1/(1 + x^4)$ centred at $c = 0$ without computing any coefficients. [/problem]

[solution] The complex extension $f(z) = 1/(1 + z^4)$ has singularities (poles) where $z^4 = -1$, i.e., $z = e^{i\pi(2k+1)/4}$ for $k = 0, 1, 2, 3$. These are the four fourth roots of $-1$: $z = e^{i\pi/4}, e^{3i\pi/4}, e^{5i\pi/4}, e^{7i\pi/4}$. Each has modulus $|z| = 1$. The distance from $c = 0$ to the nearest singularity is $1$. By the complex characterisation, $R = 1$. **Verification:** The Taylor series is $\sum_{n=0}^{\infty} (-1)^n x^{4n}$ (geometric series with ratio $-x^4$), which has $|a_{4n}| = 1$ and $a_k = 0$ for $k \not\equiv 0 \pmod{4}$. By Cauchy-Hadamard: $\limsup |a_n|^{1/n}$... taking $n = 4m$: $|a_{4m}|^{1/(4m)} = 1$, confirming $R = 1$. Despite $f$ being smooth on all of $\mathbb{R}$, the radius is finite because of the complex singularities at $|z| = 1$. $\checkmark$ [/solution]

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[problem:Uniqueness Of Coefficients From The Identity Theorem | 2] Let $f(x) = \sum_{n=0}^{\infty} a_n x^n$ and $g(x) = \sum_{n=0}^{\infty} b_n x^n$ be power series with positive radii of convergence. Suppose $f(1/k) = g(1/k)$ for all $k \in \mathbb{N}$. Prove that $a_n = b_n$ for all $n$. [/problem]

[solution] The set $S = \{1/k : k \in \mathbb{N}\}$ satisfies $f(x) = g(x)$ for all $x \in S$. The point $0$ is a limit point of $S$ (since $1/k \to 0$), and $0$ lies in the interval of convergence of both series (every power series converges at its centre). By the [Identity Theorem](/theorems/208), $f = g$ on the intersection of their intervals of convergence. In particular, $a_n = f^{(n)}(0)/n! = g^{(n)}(0)/n! = b_n$ for all $n$. This illustrates the rigidity of analytic functions: agreement on a *countable* set with a limit point forces agreement everywhere. A smooth-but-not-analytic function can agree with $0$ on all of $(-\infty, 0]$ while being nonzero on $(0, \infty)$ — the identity theorem fails precisely because such functions are not representable by convergent power series. $\checkmark$ [/solution]

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[problem:Integration Trick — Computing $\log 2$ | 3] Using the power series $\log(1 + x) = \sum_{n=1}^{\infty} (-1)^{n-1} x^n/n$ for $|x| < 1$, prove that the series also converges at $x = 1$ and that $\log 2 = \sum_{n=1}^{\infty} (-1)^{n-1}/n$. Then derive the faster-converging series $\log 2 = \sum_{n=1}^{\infty} 1/(n \cdot 2^n)$ by choosing a different power series expansion. [/problem]

[solution] **Part 1: $\log 2 = 1 - 1/2 + 1/3 - 1/4 + \cdots$.** At $x = 1$, the series $\sum (-1)^{n-1}/n$ converges by the alternating series test ($1/n$ is decreasing and tends to $0$). Abel's theorem (which states that if $\sum a_n$ converges and $f(x) = \sum a_n x^n$ for $|x| < 1$, then $\lim_{x \to 1^-} f(x) = \sum a_n$) gives: \begin{align*} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = \lim_{x \to 1^-} \log(1 + x) = \log 2. \end{align*} (Abel's theorem is needed because term-wise evaluation at boundary points is not automatic.) **Part 2: A faster series.** Start from $-\log(1 - x) = \sum_{n=1}^{\infty} x^n/n$ for $|x| < 1$. Set $x = 1/2$: \begin{align*} -\log(1/2) = \log 2 = \sum_{n=1}^{\infty} \frac{1}{n \cdot 2^n}. \end{align*} This series converges much faster: the $n$th term decays like $1/(n \cdot 2^n)$ (exponential), compared to $1/n$ (polynomial) for the alternating series. Ten terms give $\log 2 \approx 0.6931471$ to seven decimal places, while ten terms of the alternating series give only $\log 2 \approx 0.6456$ (two correct digits). The trick is choosing a point $x = 1/2$ *inside* the interval of convergence rather than at the boundary $x = 1$. Interior points give absolute convergence and exponential decay; boundary points give only conditional convergence and polynomial decay. $\checkmark$ [/solution]