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For $\sigma =\mathrm{Re}s>1$, the Dirichlet series $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ is absolutely convergent. A finite product over primes $p\le P$ can be expanded into absolutely convergent geometric series. By unique prime factorization,The omitted integers are all greater than $P$, so the difference from $\zeta (s)$ has modulus at most ${\sum }_{n>P}{n}^{-\sigma }\to 0$. ThusThis defines the infinite Euler product as the limit of finite products. Its absolute convergence also follows from ${\sum }_p{p}^{-\sigma }<\infty$, with $\log (1-p^{-s}{)}^{-1}=p^{-s}+O(p^{-2\sigma })$ for sufficiently large $p$.

The Gauss map of an oriented surface is its smooth unit normal $N:S\to S^2$. With the convention that the shape operator is $A_p=-d{N}_p$, the second fundamental form is ${\mathrm{II}}_p(v,w)=⟨A_{p}v,w⟩$ and the normal curvature in a nonzero direction $w$ is $k_n(w)=\mathrm{II}(w,w)/⟨w,w⟩$. The two principal curvatures are the real eigenvalues $k_1,k_2$ of this self-adjoint operator; the mean curvature is $H=(k_1+k_2)/2$.

For directions successively spaced around the tangent plane, write their angles to a principal direction as ${\theta }_j={\theta }_0+(j-1)\pi /m$. The Euler formula for normal curvature givesThe geometric series ${\sum }_{j=0}^{m-1}{e}^{2\pi ij/m}=0$ for $m\ge 2$ makes the cosine sum zero. ThereforeHere equal angular spacing is understood in the usual directed sense. If the printed consecutive angles were only unsigned, without requiring the same rotational sense, the assertion would be false: for $m=3$ the directions $0,\pi /3,0$ satisfy those unsigned angle conditions but do not generally give $3H$.

A minimal surface has $H=0$, so at each point $k_2=-k_1$. In an orthonormal principal basis, $A=\mathrm{diag}(k_1,-k_1)$ and $A^2=k_1^{2}I$. Hencewhich proves the identity with $\boxed{\mu (p)=k_1^2=-K(p)\ge 0}$, where $K$ is the Gaussian curvature.

The converse is false. The identity only forces $k_1^2=k_2^2$, which allows either $k_2=-k_1$ or $k_2=k_1$. A sphere of radius $R$ has $dN=I/R$ for its outward normal, so it satisfies the identity with $\mu =R^{-2}$, yet $H=-1/R\ne 0$ under the chosen convention. Thus a conformal Gauss map can also arise from an umbilic nonminimal surface.

The stage cost is $(x+u{)}^2+6u^2\ge 0$. The finite-horizon Bellman equation isInductively write $C_{s-1}(x)={\Pi }_{s-1}{x}^2$. Completing the square in $u$ gives the unique optimal first controlandFor $0\le q<2$, the recurrence map $F(q)$ satisfies $F(q)<2$ andThus ${\Pi }_s$ increases and is bounded above by two. Its limit solves $L=6(1+L)/(7+L)$, or $(L-2)(L+3)=0$; nonnegativity forces $L=2$. In particular $C_{\infty }(x_0)\ge 2x_0^2$, and indeed this recurrence already identifies the limit.

For the proposed policy, induction on the state equation gives $x_t=(2/3{)}^t{x}_0$ and $u_t=-x_t/3$. The stage cost is $(10/9)x_t^2$, so the infinite geometric series givesThis supplies an admissible policy attaining the lower bound:The last formula is the requested closed-loop control: it uses the observed current state rather than an initial-state schedule.

Let $K=\{z:|\mathrm{Re}z|+|\mathrm{Im}z|\le 1\}$. Choose a unit complex number $u$ normal to a supporting side of this convex set, withSuch a choice exists explicitly: take $u=(s+it)/\sqrt{2}$, where $s,t\in \{-1,1\}$ are the signs of the real and imaginary parts of $\zeta$; then $h=1/\sqrt{2}$ and the strict inequality is exactly the hypothesis. Set $c=-Ru$. Since $|z|\le 1$ on $K$,For sufficiently large $R$, the former is strictly smaller than the latter uniformly on $K$. Consequently $q={\max }_K|z-c|/|\zeta -c|<1$.

The geometric series now supplies actual approximating polynomials:Choose $N$ so the right-hand side is below $ϵ$. This proves the required uniform approximation without assuming a general approximation theorem.

For a series $\sum u_n$ with $u_n>0$, suppose $L={\lim }_{n\to \infty }{u}_{n+1}/u_n$ exists, allowing $L=+\infty$. The ratio test says the series converges if $L<1$ and diverges if $L>1$. In the first case an eventual ratio bound by some $q<1$ compares the tail to a geometric series. In the second case an eventual ratio above some $q>1$ prevents the terms tending to zero.

If $L=1$, the test is inconclusive: both $\sum 1/n$ and $\sum 1/n^2$ have limiting ratio one, while the first diverges and the second converges. If the ratio has no limit, this particular limit formulation gives no conclusion.

The geometric family law on the nonnegative integers has probability generating functionIts expected value is $G^{\prime }(1)=1$, so this is a geometric-offspring critical branching process. Starting from $G_0(s)=s$, verify the rational iterate by induction. If $G_n(s)=[n-(n-1)s]/[n+1-ns]$, thenwhich has the same form with $n$ replaced by $n+1$. Thus, for $n\ge 1$,The stated disk is the actual power-series convergence domain, not merely a place where the rational continuation can be evaluated. Its simple pole is at $s=1+1/n$.

The extinction probability at generation $n$ is the constant coefficient:For later use, expanding the remaining geometric series givesIn particular $P(X_n>0)=1/(n+1)$. At $n=0$ the initial population is one deterministically; its PGF is $s$, and the displayed radius involving $1/n$ is used only for $n\ge 1$.

For events with $P(B)>0$, the conditional probability is $P(A\mid B)=P(A\cap B)/P(B)$. Applying the definition in both orders gives Bayes' theorem:Conditionally on $N=n$, independent fair tosses give a binomial distribution for the head count, so $P(H=1\mid N=n)=n{2}^{-n}$. The joint probability mass function is consequentlyUse the law of total probability and the differentiated geometric series, ${\sum }_{n\ge 1}n{r}^n=r/(1-r{)}^2$, to obtain $P(H=1)=4/9$. The posterior coin count after exactly one head is thereforeThese probabilities sum to one by the same series identity. The conditioning weights possible coin counts by their likelihood of producing exactly one head.

For the causal case $|\varphi |<1$, expandHence the white noise coefficients are ${\psi }_0=1$ and ${\psi }_j=(\varphi +\theta ){\varphi }^{j-1}$ for $j\ge 1$. Orthogonality of white noise gives ${\gamma }_k={\sigma }^2{\sum }_{j\ge 0}{\psi }_j{\psi }_{j+k}$ for $k\ge 0$. Summing the geometric series yields the autocovariance of a causal ARMA(1,1) process:For $\varphi =0$, this means ${\gamma }_1={\sigma }^2\theta$ and ${\gamma }_k=0$ for $k\ge 2$, together with ${\gamma }_0={\sigma }^2(1+{\theta }^2)$. For $\theta =-\varphi$, all positive-lag covariances vanish and ${\gamma }_0={\sigma }^2$, as expected from cancellation of the two filter polynomials.

The root $\omega$ must be a primitive $n$th root of unity, for example $e^{2\pi i/n}$. The printed statement merely says $n$th root; that is insufficient. With $n=2$ and $\omega =1$, the states with $r=0$ and $r=1$ coincide. More generally, a root of order $d<n$ repeats the $r$ labels modulo $d$. For $n=1$ the construction is trivial.

For a primitive root of unity, the generalized Bell basis obeysThe last equality follows from the finite geometric series: a nonzero exponent difference modulo $n$ has sum zero, while zero difference has sum $n$. There are $n^2$ orthonormal vectors in an $n^2$-dimensional Hilbert space, so they form a complete basis.

For qudit teleportation, Alice's input is $|\chi {⟩}_U={\sum }_j{c}_j|j⟩$. Alice and Bob share $|{\psi }_{00}{⟩}_{aB}=n^{-1/2}{\sum }_k|kk⟩$. Define the qudit shift and phase operators by $X|j⟩=|j+1⟩$ and $Z|j⟩={\omega }^j|j⟩$. Alice measures $Ua$ in the generalized Bell basis. On outcome $(r,s)$, Bob's unnormalized state isAll outcomes have probability $1/n^2$. Alice communicates $(r,s)$ and Bob applieswhich restores $|\chi ⟩$ exactly. The protocol consumes one maximally entangled pair of $n$-level systems, a local $n^2$-outcome measurement, and a classical message with $n^2$ possibilities. For a fixed-length binary encoding, $\lceil {\log }_2{n}^2\rceil$ bits suffice. No measurement depends on the unknown amplitudes.

The integer endpoint hypothesis is important. Let $r_{n,k}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)f(k/n)-\lfloor \left(\genfrac{}{}{0ex}{}{n}{k}\right)f(k/n)\rfloor$. The floor function gives $0\le r_{n,k}<1$, while $r_{n,0}=r_{n,n}=0$ because $f(0)$ and $f(1)$ are integers. Therefore the rounded Bernstein polynomial satisfiesFor $0\le x\le 1/4$, the successive terms have ratio $x/(1-x)\le 1/3$, so the geometric series estimate givesThe last inequality follows by maximizing $x(1-x{)}^{n-1}$ at $x=1/n$. The same bound holds on $[3/4,1]$ by reflection. On $[1/4,3/4]$, each factor $x$ and $1-x$ is at most $3/4$, giving $S_n(x)\le (n-1)(3/4{)}^n$. ThusHere $n\ge 2$; for $n=1$ the difference vanishes. Without the integer endpoint hypothesis the claim would fail: at $x=0$ the error is $f(0)-\lfloor f(0)\rfloor$, independently of $n$.

The geometric series formula and part (b) givePut $\alpha =\log 3/\log 5$. Then $5^{\alpha }=3$, so $n^{\alpha }<3^{m+1}$ andThis is the largest possible decay exponent: at $n=5^m$, an estimate with exponent $\beta >\alpha$ would require $\frac{1}{2}\le c(3/5^{\beta }{)}^m$, whose right side tends to zero. Any smaller exponent also gives a valid weaker estimate. In fact $\frac{1}{2}{n}^{-\alpha }\le E_n(f)<\frac{3}{2}{n}^{-\alpha }$ for every $n\ge 1$.

The printed interval has a typo: its right endpoint must be $2/3+\delta$, rather than $1/3+\delta$. Keep $\varphi$ as an unnormalized indicator function; only $\psi$ is normalized to have integral one. Thus useWith the Boolean literal printed interval, the convolution vanishes at $2/3$, so a sequence consisting entirely of $2/3$ is already a counterexample to the proposed majorization. For the corrected interval and every $t\in [1/3,2/3]$, all $t-s$ with $|s|\le \delta$ belong to the support of $\varphi$. Hence $f(t)=1$ there. Everywhere $0\le f\le 1$, giving $S_N\le {\sum }_{n\le N}f(u_n)$, including interval endpoints.

Put $L=1/3+2\delta$. For $k\ne 0$, direct integration and the convolution theorem give the exact Fourier coefficientsIn particular,This also supplies the required justification of the Fourier series. The convolution $f$ is continuous: it is the overlap length of two moving intervals, divided by $2\delta$. Its Fourier coefficients are absolutely summable, so the corresponding Fourier series converges uniformly to a continuous function $g$. Termwise integration gives $\hat{g}(k)=\hat{f}(k)$; the permitted uniqueness assertion applied to $f-g$ yields $g=f$.

Let $A_k={\sum }_{n\le N}{e}^{2\pi ik{u}_n}$. Split the absolutely convergent Fourier series at the arbitrary real cutoff $M>1$:There are fewer than $2M$ low-frequency terms and each coefficient has magnitude at most $L<1$. For the tail, $|A_k|\le N$ and ${\sum }_{|k|\ge M}{k}^{-2}\le C/M$. ThereforeThis proves the requested interval discrepancy estimate without an unjustified sharp truncation of the Fourier series.

For a nonzero integer $b$, let $a$ be the nearest integer to $b\sqrt{2}$. The integer $a^2-2b^2$ is nonzero, and $|a|\le \sqrt{2}|b|+1/2$. The suggested factorization yieldsThis is stronger than the stated lower bound for the fractional part, since the distance to the nearest integer is at most that fractional part. For $u_n=n\sqrt{2}$, the geometric series and $|\sin (\pi t)|\ge 2\Vert t{\Vert }_{\mathbb{R}/\mathbb{Z}}$ giveThus the low-frequency contribution is $O(M^2)$. Choose $M=\lceil 100N^{2/5}\rceil$ and $\delta =(2\sqrt{M}{)}^{-1}$; these choices satisfy $0<\delta <1/10$ for every $N\ge 1$. Both $\delta N$ and $N/(\delta M)$ are $O(N^{4/5})$, and so is $M^2$. Hence

Put $d=np=\gamma \log n$ and $\mu =n^{1-\gamma }$. The hypothesis implies $\mu \ge e^{\omega (n)}\to \infty$. For the number $I$ of isolated vertices, the isolated vertices in the Erdős-Rényi model formulas giveThus the Chebyshev inequality gives $I=(1+o(1))\mu$ with high probability, uniformly over the specified $\gamma$ range.

The Cayley formula supplies a spanning tree on any connected vertex set. Consequently the expected value $C_j$ of the number of graph components of order $j\ge 2$ is bounded byChoose a fixed $\eta \in (0,1/2)$ with ${\gamma }_0(1-\eta )(k+1)>1$. For $k+1\le j\le \eta n$, with $j\ge 2$, the binomial coefficient bound $\left(\genfrac{}{}{0ex}{}{n}{j}\right)\le (en/j{)}^j$ givesHere $ed{e}^{-d(1-\eta )}\le e\log n{n}^{-{\gamma }_0(1-\eta )}\to 0$. The geometric series and a union bound therefore exclude this entire size range, since $n[e\log n{n}^{-{\gamma }_0(1-\eta )}{]}^{k+1}=o(1)$. For $\eta n\le j\le n/2$, an empty graph cut would be necessary. Its total probability is at mostThere are consequently no graph components of orders between $k+1$ and $n/2$, with high probability.

For each fixed $2\le j\le k$, a graph component containing a graph cycle has a spanning tree and at least one extra edge. Counting that extra edge gives an expected value $O_k(d^j{e}^{-dj})=o(1)$; hence every small graph component is a tree component. The total number $R_2$ of vertices in tree components of orders $2,\dots ,k$ satisfiesThe Markov inequality gives $R_2=o(\mu )$ with high probability. When $k=1$, this sum is empty and $R_2=0$. Since $\mu =o(n)$, the remaining vertices must lie in one graph component larger than $n/2$. It is unique, and its order is $n-(1+o(1))n^{1-\gamma }$. Every other graph component is a tree with at most $k$ vertices. This is the logarithmic-regime giant component mechanism: almost all missing vertices are isolated.

Every nonprincipal Dirichlet character modulo the prime $q$ is primitive, so its finite Fourier coefficients have modulus one off zero; the coefficient at zero is zero by character orthogonality. Fourier inversion theorem givesThe finite geometric series givesThe printed hint omits $n$ from the exponential; its literal constant summand would not obey the bound for arbitrary $N$. The geometric-series calculation proves the needed estimate independently. Pairing $a$ with $q-a$ givesThe last sum is a harmonic number. This proves the Pólya–Vinogradov inequality uniformly in $M$ and $N$; complete blocks of length $q$ also vanish by Orthogonality of Dirichlet characters.

Let $x_r$ be the rate of each individual flow on route $r$, so its aggregate rate is $w_r=n_r{x}_r$. The feasible allocations satisfy ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$. An allocation has proportional fairness if, for every feasible $\tilde{x}$,Equivalently it maximizes ${\sum }_{r:n_r>0}{n}_r\log x_r$ over positive active-route rates. This is the first-order optimality condition for a concave objective on a convex set. Rates for absent flows can be set to zero.

For the linear flow network, put $N={\sum }_{r=0}^I{n}_r$ and $M={\sum }_{i=1}^I{n}_i$. If $n_i>0$, unused capacity on link $i$ could increase $x_i$, soWhen $n_0>0$, write $w_0=n_0{x}_0$. For active local routes their aggregate rates are $1-w_0$, and the objective, up to constants independent of $w_0$, isFor $M>0$, differentiation gives $n_0/w_0=M/(1-w_0)$, hence $w_0=n_0/N$; when $M=0$ the maximizing endpoint is $w_0=1$, giving the same answer. Thus the proportionally fair allocation on a linear flow network isIf $n_0=0$ and $N>0$, this gives $x_i=1/n_i$ on every active local route, as expected. No departure occurs at the empty state.

Independent Poisson processes of flow arrivals and independent exponential distributions of document sizes make the count vector a continuous-time Markov chain. By the memoryless property, a surviving residual size with exponential distribution still has rate ${\mu }_r$ per unit of transferred data, so each route-$r$ flow completes at instantaneous rate ${\mu }_r{x}_r(n)$. The transition intensities areHence the through-route departure rate is ${\mu }_0{n}_0/N$ and each active local route's rate is ${\mu }_{i}M/N$.

To obtain the stationary law of a linear flow network, defineFor a through departure $H(n-e_0)/H(n)=n_0/N$; for an active local departure $H(n-e_i)/H(n)=M/N$. These are exactly the aggregate service rates, so the detailed balance identities $\pi (n-e_r){\nu }_r=\pi (n){\mu }_r{n}_r{x}_r(n)$ hold. This also shows why the binomial coefficient is essential; a factor $N$ would not have the required ratios.

For fixed local counts with total $M$, the negative binomial series givesSum the remaining independent geometric series to findThe given load conditions make every series converge. Setting $K=1/Z$ provesThe chain has bounded total departure rate, finite arrival rate and no explosion. With positive arrival rates it is irreducible, and this normalized invariant law gives the positive recurrent Markov chain property; zero arrival rates restrict the stationary support to the corresponding reachable class.

Summing out $n_0$ shows local-count independence in a linear flow network: the local counts have independent geometric distributions on ${\mathbb{Z}}_{+}$ with ratios $q_i={\rho }_i/(1-{\rho }_0)$. ThereforeThey are independent of one another under the stationary marginal, despite sharing the random through-flow population. The through count is generally dependent on them.

The PDF's small superscript is essential: $f(x)=a^x\mathrm{mod}N$. Interpret $r$ as the least positive period, the multiplicative order of $a$ modulo $N$. If one permitted an unspecified nonminimal period, the answer would not be identifiable; for example, $a=1,N=8$ gives the same constant function with period one or period eight. Under the promised periodic modular-exponential interpretation, $a^r\equiv 1(\mathrm{mod}N)$, so $a$ is a unit. ThusIn particular, different residue classes modulo $r$ have different function values, which is necessary for the usual quantum period finding coset argument.

Prepare $N^{-1/2}{\sum }_x|x⟩|1⟩$ by applying the quantum Fourier transform to the first register. Use repeated squaring to precompute $a^{2^j}\mathrm{mod}N$, then multiply the value register by that known number controlled on bit $j$ of $x$. These controlled modular multiplications are reversible because $a$ is a unit; their inverses use the modular inverses of the same constants. This produces $N^{-1/2}{\sum }_x|x⟩|a^x\mathrm{mod}N⟩$ with a polynomial number of the assumed arithmetic operations. Known work registers can be uncomputed.

Measure the value register. Since $r\mid N$, its fiber is exactly one coset, and the first register becomesThe quantum Fourier transform of a periodic coset state is supported uniformly on the $r$ outcomes $c=jN/r$, $0\le j<r$. Indeed the inner geometric series ${\sum }_{\ell =0}^{N/r-1}{e}^{2\pi i\ell rc/N}$ is zero unless $(N/r)\mid c$, and each allowed amplitude has modulus $1/\sqrt{r}$.

Take two independent Fourier samples $c_1,c_2$ and returnThis is two-sample exact period recovery. The Chinese remainder theorem makes divisibility of independent uniform residues independent across the different prime divisors of $r$. For each such prime $\ell$, both $j_1$ and $j_2$ are divisible by $\ell$ with probability ${\ell }^{-2}$. ThereforeThe number-theoretic facts used here are the Chinese remainder theorem, the Euler product for the Riemann zeta function at two, and $\zeta (2)={\pi }^2/6$. The candidate can also be certified by $a^{\hat{r}}\equiv 1(\mathrm{mod}N)$: it divides the least period, so this equality holds exactly when it is the full period. Each preparation, reversible evaluation, QFT, measurement and greatest common divisor calculation has the assumed or standard polynomial time cost in $\log N$. Two runs suffice for the desired constant success probability. Quantum period finding recovers the least period with probability at least $6/{\pi }^2$ in polynomial time.

Let $r=e^{i\theta }\ne 1$. The geometric series from any starting index is bounded:Using summation by parts,Since the coefficients are nonincreasing and positive, the modulus is at most $K(a_N+a_m-a_N)=K{a}_m\to 0$. The Cauchy criterion proves convergence for every $\theta \notin 2\pi \mathbb{Z}$. This is a direct proof of the Dirichlet test in this setting.

For $a_n=1/n$, the imaginary parts of the convergent partial sums give convergence of the sine series at those angles. If $\theta \in 2\pi \mathbb{Z}$, every sine term is zero. Hence the sine series converges for all real angles.

A real sequence $(x_n)$ has limit of a sequence $x$ whenA series ${\sum }_{n\ge 1}{x}_n$ is a convergent series with sum $s$ when its partial sums $S_N={\sum }_{n=1}^N{x}_n$ have limit of a sequence $s$.

For the first comparison test for series, either of the two allowed bounds implies $0<x_n\le a_n+b_n$. HenceThe partial sums form a monotone bounded sequence, so they converge. The series $\sum x_n$ is convergent.

For the exponent $2$, use $1/n^2\le 1/[n(n-1)]=1/(n-1)-1/n$ for $n\ge 2$. The resulting telescoping series bounds every partial sum by $2$, proving that the series converges by the monotone bounded sequence property. To prove divergence of the harmonic series, group the terms with $2^j<n\le 2^{j+1}$: their sum is at least $2^j/2^{j+1}=1/2$. Thus its partial sums are unbounded. If $\alpha \le 1$, then $n^{-\alpha }\ge n^{-1}$, so the comparison test for series proves

For the weighted square assumption, the finite-dimensional Cauchy-Schwarz inequality gives, for every $N$,Again the positive partial sums form a monotone bounded sequence. This is the weighted square roots of a summable sequence argument, with $a_n=n^2{x}_n^2$. The weighted square condition implies $\sum x_n<\infty$.