Sequences & Series

Differentiate the geometric series: $\sum nx^{n-1} = \tfrac{1}{(1-x)^2}$, so $\sum\tfrac{n}{2^n} = 2$ and $\sum\tfrac{n+1}{3^n} = \tfrac94$. Integrate it: $\sum\tfrac{x^n}{n}$, so $\sum\tfrac{1}{n2^n} = \ln 2$. Add the classical constants $\zeta(2) = \tfrac{\pi^2}{6}$, $\eta(2) = \tfrac{\pi^2}{12}$, $\sum\tfrac{(-1)^{n+1}}{n} = \ln 2$, $\sum\tfrac{(-1)^n}{2n+1} = \tfrac\pi4$. A Putnam series problem is usually one of these wearing a costume.

$\sum\frac{1}{n(n+1)} = 1$; $\sum\frac{1}{n(n+2)} = \tfrac12(1 + \tfrac12) = \tfrac34$ (lag $2$: two front terms survive); $\sum\frac{1}{n(n+1)(n+2)} = \tfrac12\cdot\tfrac{1}{1\cdot2} = \tfrac14$. Products telescope too: $\prod\left(1 - \tfrac{1}{n^2}\right) = \prod\tfrac{(n-1)(n+1)}{n^2} \to \tfrac12$. The tail must go to $0$ — say so.

Stolz–Cesàro is L'Hôpital for sequences: it turns a ratio of sums into a ratio of terms. $\dfrac{\sqrt1 + \cdots + \sqrt n}{n^{3/2}} \to \dfrac{\sqrt n}{\tfrac32\sqrt n} = \tfrac23$ — the same answer the Riemann sum $\int_0^1\sqrt x\,dx$ gives. Sums of $f(k/n)$ over $k$, divided by $n$, are integrals; sums $\sum_{k=1}^n\frac{1}{n+k} = \frac1n\sum\frac{1}{1 + k/n} \to \ln 2$.

$\tfrac13\left(\tfrac1n - \tfrac{1}{n+3}\right)$ telescopes with lag $3$: the surviving front terms are $1 + \tfrac12 + \tfrac13 = \tfrac{11}{6}$, so the sum is $\tfrac13\cdot\tfrac{11}{6} = \tfrac{11}{18}$.

Fix $\varepsilon > 0$; for $n \ge N$, $(L - \varepsilon)(b_{n+1} - b_n) < a_{n+1} - a_n < (L + \varepsilon)(b_{n+1} - b_n)$. Sum these from $N$ to $n - 1$ (telescoping!): $(L - \varepsilon)(b_n - b_N) < a_n - a_N < (L + \varepsilon)(b_n - b_N)$. Divide by $b_n \to \infty$: the terms $a_N/b_n$ and $b_N/b_n$ vanish, leaving $L - \varepsilon \le \liminf\tfrac{a_n}{b_n} \le \limsup\tfrac{a_n}{b_n} \le L + \varepsilon$. $\blacksquare$ Significance: it is the discrete mean value theorem, and it proves Cesàro means preserve limits ($b_n = n$) — a tool for every "average of a sequence" problem.

$\zeta(2) = \sum\frac{1}{n^2} = \sum\int_0^1\!\!\int_0^1(xy)^{n-1}dx\,dy = \iint_{[0,1]^2}\frac{dx\,dy}{1 - xy}$ (geometric series, swap by monotone convergence). Rotate coordinates $45^\circ$ ($u = \tfrac{x+y}{\sqrt2}$, $v = \tfrac{y - x}{\sqrt2}$); the integrand becomes $\tfrac{1}{1 - \tfrac{u^2 - v^2}{2}}$ and the inner integral is an arctangent; two elementary trigonometric integrals later the total is $\tfrac{\pi^2}{6}$ (Beukers–Calabi–Kolk). $\blacksquare$ The lesson for the exam: sums become integrals ($\tfrac{1}{n} = \int_0^1x^{n-1}dx$) and integrals become sums; move freely between them.

Step 1 — from $\sum x^n = \tfrac{1}{1-x}$: differentiate and multiply by $x$ twice. $\sum nx^n = \tfrac{x}{(1-x)^2}$; then $\sum n^2x^n = x\tfrac{d}{dx}\tfrac{x}{(1-x)^2} = \tfrac{x(1 + x)}{(1-x)^3}$.

Step 2 — at $x = \tfrac12$: $\tfrac{\tfrac12\cdot\tfrac32}{\tfrac18} = 6$.

Step 3 — check the first terms: $\tfrac12 + 1 + \tfrac98 + 1 + \tfrac{25}{32} + \cdots \approx 4.4$ after five terms, climbing toward $6$ ✓. The operator "$x\frac{d}{dx}$" multiplies coefficients by $n$; apply it as many times as the power of $n$ demands.

Step 1 — the terms decrease, so Cauchy condensation applies: the series converges iff $\sum 2^k\cdot\frac{1}{2^k(\ln 2^k)^p} = \sum\frac{1}{(k\ln 2)^p}$ does.

Step 2 — that is a $p$-series in $k$: converges iff $p > 1$.

Step 3 — so $\sum\frac{1}{n\ln n}$ diverges (barely!), $\sum\frac{1}{n(\ln n)^2}$ converges. The integral test says the same: $\int\frac{dx}{x(\ln x)^p}$ with $u = \ln x$ becomes $\int\frac{du}{u^p}$. Logarithmic factors change convergence only at the razor's edge $p = 1$.