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Put $s=\sin x$, so $-1\le s\le 1$. The printed summand is $(3s^n+s^{2n})/n$. If $|s|<1$, its absolute value is at most $3|s{|}^n+|s{|}^{2n}$, and the two geometric series converge. Thus there is absolute convergence.

For $s=1$ the series becomes $\sum 4/n$, a divergent multiple of the harmonic series. For $s=-1$ it becomesThe first series converges by the alternating series test, while the second diverges to $+\infty$, so their partial sums tend to $+\infty$. In particular, alternation of some terms does not imply convergence here. The complete classification isThere are no conditionally convergent cases.

Let $r=|\sin x|\in [0,1]$. For $r<1$, the absolute summand $r^n/\sqrt{n}$ is at most $r^n$, so a geometric series comparison proves absolute convergence. For $r=1$, the series is $\sum (-1{)}^n/\sqrt{n}$: its positive term sizes decrease to zero, so the alternating series test proves convergence. Its absolute series is a divergent p-series with exponent $1/2$. HenceIt converges for every real $x$ and never diverges.

For fixed real $x$, define $u_0=x$ and $u_n=\sin (0.99u_{n-1})$. This makes $u_n$ exactly the summand containing $n$ nested sines. The elementary inequality $|\sin t|\le |t|$ gives inductivelyThe fact that an iterated contraction gives a summable orbit applies here. Comparison with a geometric series therefore givesThus the nested-sine series converges absolutely for every real $x$; there are no divergent or merely conditionally convergent cases.

The contraction mapping theorem states that a contraction mapping $T:X\to X$ on a nonempty complete metric space has a unique fixed point. Here contraction means that some $0\le q<1$ satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$.

For a proof, choose $x_0$ and iterate $x_{r+1}=T{x}_r$. Induction gives $d(x_{r+1},x_r)\le q^{r}d(x_1,x_0)$. For $m>n$, the triangle inequality and a geometric series giveThus the iterates form a Cauchy sequence, and completeness supplies a limit $a\in X$. A contraction mapping is continuous, so $Ta=\lim T{x}_n=\lim x_{n+1}=a$. If $b$ is another fixed point, $d(a,b)=d(Ta,Tb)\le qd(a,b)$, forcing $d(a,b)=0$. This proves existence and uniqueness.

For the exponential problem, use $h=f\circ f$ on the complete metric space $\mathbb{R}$. Its derivative isPutting $t=e^{-x}>0$, the function $t{e}^{-t}$ has maximum $1/e$ at $t=1$, as its derivative is $(1-t)e^{-t}$. By the mean value theorem, $|h(x)-h(y)|\le e^{-1}|x-y|$, so $h$ is a contraction mapping on all of $\mathbb{R}$. Let $a$ be its unique fixed point. The compositions commute, so $h(f(a))=f(h(a))=f(a)$; uniqueness of the fixed point of $h$ forces $f(a)=a$. Conversely every fixed point of $f$ is fixed by $h$. Hence there is exactly one real solution of $x=e^{-x}$. It lies in $(0,1)$ because $a=e^{-a}>0$ and then $e^{-a}<1$. This uses the iterated contraction argument without claiming that $f$ itself contracts on all of $\mathbb{R}$.

On any compact set avoiding the integers, writeFor all sufficiently large $|n|$, uniformly on that compact set this is $O(|n{|}^{-2})$. The summands in the higher series are $O(|n{|}^{-k})$ for $k\ge 2$. The Weierstrass M-test therefore gives locally uniform convergence away from the integer poles and permits termwise differentiation. Near an integer $m$, isolate its singular summand; all the others converge uniformly on a small neighborhood. Thus the first function has a simple pole of residue one at each integer, and the $k$th function has a pole of order $k$ there with principal part $(z-m{)}^{-k}$. There are no other singularities, so all are meromorphic functions.

Pairing positive and negative indices is legitimate because the regularized first series converges absolutely. It becomesHere is a direct derivation of its cotangent partial-fraction expansion. Fix $z\notin \mathbb{Z}$ and integrate $F(\zeta )=\pi \cot (\pi \zeta )/({\zeta }^2-z^2)$ around the square with real and imaginary coordinates $\pm R$, $R=N+1/2$, taking $N$ large enough to enclose $\pm z$. On the vertical sides $|\cot \pi \zeta |\le 1$, and on the horizontal sides it is at most $\coth (\pi R)$. Thus the cotangent is uniformly bounded and $|{\zeta }^2-z^2|\ge R^2-|z{|}^2$, so the contour integral is $O(R^{-1})\to 0$.

The residues at integers are $1/(n^2-z^2)$; the two residues at $z$ and $-z$ sum to $\pi \cot (\pi z)/z$. The residue theorem and absolute convergence of the integer residue sum consequently giveUsing the exponential definitions of sine and cosine now yields, for $w=e^{2\pi iz}$,The same locally uniform convergence gives ${\epsilon }_k^{\prime }=-k{\epsilon }_{k+1}$, including $k=1$. Hence if ${\epsilon }_k(z)={\phi }_k(e^{2\pi iz})$, the chain rule givesThis is a rational function whenever ${\phi }_k$ is, proving the requested descent for every $k$ by induction, withIn particular periodicity is a consequence of the explicit descent, rather than an assumption about the summation order.

Finally, for $|z|<1$ expand each paired summand as a geometric series:On $|z|\le r<1$ the double series converges absolutely, since its absolute sum is at most $2r{\sum }_{n\ge 1}{n}^{-2}/(1-r^2)$. Interchanging the sums and using the Riemann zeta function gives the Laurent seriesThus the $z^{-1}$ coefficient is one; the positive odd coefficients are $-2\zeta (2j+2)$, and all other coefficients vanish.

Put $e_N(t)=e^{2\pi it/N}$ and use normalized averages ${\mathbb{E}}_x=N^{-1}{\sum }_{x\in \mathbb{Z}/N\mathbb{Z}}$. Our convention for Fourier coefficients on a finite abelian group isThe underlying orthogonality of complex exponentials is ${\mathbb{E}}_x{e}_N(rx)=1$ for $r=0$ and zero otherwise. For nonzero $r$ this is the finite geometric series with ratio $e_N(r)\ne 1$; its sum vanishes because that ratio has $N$th power one. The identical orthogonality statement holds when averaging over $r$ instead of $x$.

Using this orthogonality to sum the definition over $r$ proves Fourier inversion on a finite group:Indeed $\mathbb{E}f\overline{g}={\sum }_r\overline{\hat{g}(r)}{\mathbb{E}}_{x}f(x)e_N(-rx)={\sum }_r\hat{f}(r)\overline{\hat{g}(r)}$. This proves the Parseval identity on a finite group and its squared-norm case:For normalized convolution on a finite group, write $(f\ast g)(x)={\mathbb{E}}_{y}f(y)g(x-y)$. Changing variables $x=y+z$ in its transform givesthe convolution theorem on a finite group. Linearity of the transform follows directly from its definition. Translation ${\tau }_{t}f(x)=f(x+t)$ gives $\widehat{{\tau }_{t}f}(r)=e_N(rt)\hat{f}(r)$ by changing variables. Complex conjugation gives $\hat{\overline{f}}(r)=\overline{\hat{f}(-r)}$; for $\tilde{f}(x)=\overline{f(-x)}$, the corresponding identity is $\hat{\tilde{f}}(r)=\overline{\hat{f}(r)}$. Finally $\hat{f}(0)=\mathbb{E}f$ and $|\hat{f}(r)|\le \mathbb{E}|f|$ by the triangle inequality. These identities establish the basic normalized Fourier analysis on a finite abelian group needed below, with all normalizations fixed.

For completeness, pointwise multiplication has the dual convolution formula $\widehat{fg}(r)={\sum }_s\hat{f}(s)\hat{g}(r-s)$: insert the inversion formula for $g$ into the defining average and collect its frequencies. If $c\ne 0$ modulo $N$, the change of variables $u=cx$ gives $\widehat{f(c\cdot )}(r)=\hat{f}(r{c}^{-1})$. These also follow directly from the same additive character orthogonality.

Write $e(t)=e^{2\pi it}$. We first prove the needed cancellation of quadratic exponential sums, rather than invoke equidistribution without proof. The number $\sqrt{2}$ is irrational: writing it as a reduced fraction would make its numerator even from $u^2=2v^2$, and then its denominator even as well, a contradiction. Thus $2hr\sqrt{2}$ is irrational for nonzero integers $h,r$.

Fix $h\ne 0$ and set $z_n=e(h\sqrt{2}{n}^2)$ for $1\le n\le L$, extending it by zero otherwise. For integers $L\ge H\ge 1$, each term appears exactly $H$ times in the shifted sum, soFor any complex numbers $w_1,\dots ,w_J$, nonnegativity of ${\sum }_j|w_j-J^{-1}{\sum }_i{w}_i{|}^2$ gives $|{\sum }_j{w}_j{|}^2\le J{\sum }_j|w_j{|}^2$. Apply this special Cauchy-Schwarz inequality to the displayed outer sum, and then expand the inner squared modulus. This givesFor fixed $H$ and large $L$, each correlation is the geometric serieswhose absolute value is at most $2/|1-e(2hr\sqrt{2})|$, independently of $L$. This follows by multiplying the geometric sum by $1-e(2hr\sqrt{2})$; its denominator is nonzero by irrationality. Divide the inequality by $L^2$ and let $L\to \infty$, keeping $H$ fixed. It gives a limit superior at most $1/H$. Taking arbitrarily large $H$ provesThis is the quadratic instance of the Van der Corput inequality for finite scalar sequences, with its proof supplied here.

To turn cancellation into the particular approximation, put $\epsilon =10^{-8}$ and choose an integer $M>2/{\sin }^2(\pi \epsilon )$. The Fejér kernel is the explicit finite trigonometric polynomialThe expansion follows by counting pairs of indices with difference $h$. Its constant coefficient is one, so the cancellation just proved givesIf $\Vert t{\Vert }_{\mathbb{R}/\mathbb{Z}}\ge \epsilon$, the geometric-series formula givesFor all sufficiently large $L$, the displayed average exceeds $1/2$. Consequently at least one $n\in \{1,\dots ,L\}$ satisfiesIt is a positive integer, as required. Every auxiliary cancellation and kernel identity used in this existence proof has been derived; a smooth cutoff or an unproved Weyl criterion is unnecessary.

Set $q=1-p<1/2$. Declare a bond of the planar dual graph open when its crossing primal bond is closed. These states have the independent dual bond percolation law with parameter $q$; the dual square lattice is a translate of the primal square lattice. Part (i) supplies an $a=a(q)>0$ such thatIn particular every dual cluster is finite almost surely.

Suppose the primal percolation cluster $C_0$ is finite. Fill its finite complementary holes. Every bond on the resulting outer boundary is closed, since an open bond leaving the cluster would reach another cluster vertex. The crossing dual bonds form a connected enclosing contour, possibly touching itself, contained in one dual cluster $K$. This is the dual contour bound for a finite planar cluster. If $m=|K|$, a connected set of $m$ square-lattice dual vertices has each coordinate width at most $m-1$. All vertices of $C_0$ lie inside the enclosing contour, soMoreover the contour surrounds $0$ and the diameter of $K$ is at most $m-1$ in the graph distance. Thus every vertex of $K$ lies in the square of radius $m+1$ about $0$. There are at most $(2m+3{)}^2$ possible dual vertices in that square, with a harmless enlargement if a boundary convention requires it.

On $\{n\le |C_0|<\infty \}$ there is therefore an integer $m\ge \lceil \sqrt{n}\rceil$ and a dual vertex in that square whose dual cluster has size exactly $m$. The enclosing contour has at least four vertices. A union bound and part (i) givefor constants $C_1,c_1>0$ depending only on $q$. The last estimate follows by bounding the polynomial factor by a constant times $e^{am/2}$ and summing a geometric series. This explains the square root: an enclosing dual cluster must have size at least the square root of the primal volume.

For $p<1$, the Harris-Kesten theorem and $p>p_c$ imply $\theta (p)>0$, so $v:={\mathbb{P}}_p(|C_0|<\infty )=1-\theta (p)<1$. Choose $N$ with $\log C_1\le (c_1/2)\sqrt{N}$. For $n\ge N$ use the preceding bound; for $1\le n\le N$ use the bound $v$. The positive constant $b=\min \{c_1/2,-\log v/\sqrt{N}\}$ gives the required conclusion:For $p=1$ the event is empty, so any $b>0$ works.

Put $\mu ={\mu }_1+{\mu }_2$. In this heterogeneous two-server queue, the total population alone is not a Markov process: when only one customer is present, its departure rate depends on which server is occupied. Use states $0,a,b,2,3,\dots$, where $a$ and $b$ distinguish the two single-customer possibilities, and $n\ge 2$ records the total population. The memoryless property of the exponential distributions gives these nonzero transition rates:and $q_{n,n+1}=\nu$ for $n\ge 2$, $q_{n,n-1}=\mu$ for $n\ge 3$. In particular, a departure from state $2$ leaves the other server occupied.

The detailed balance equations suggestFor example, both ${\pi }_a\nu ={\pi }_2{\mu }_2$ and ${\pi }_b\nu ={\pi }_2{\mu }_1$ hold; these are the two transitions that would be lost by using only population as the state. All other adjacent-state detailed balance equations hold as well. Since $\nu /\mu <1$, the geometric series converges, and normalization givesThus the stationary distribution exists, and the requested probability isWith equal server rates ${\mu }_1={\mu }_2=m$, this reduces to ${\nu }^2/[m(2m+\nu )]$, as for an M-M-s queue with two servers.

To prove the output claim, put the continuous-time Markov chain in its stationary distribution. The preceding detailed balance equations make it a reversible Markov chain: time reversal of a continuous-time Markov chain has the same transition rates. A forward departure becomes a reversed arrival. In every reversed state, the total rate of such arrivals is $\nu$: from $0$ the rates $\nu /2$ add, from $a$ or $b$ the upward rate is $\nu$, and from every $n\ge 2$ it is also $\nu$. The reversed process can therefore be constructed with an independent rate-$\nu$ Poisson process of arrival epochs, choosing the idle server by a fair coin when appropriate. Consequently reversed arrival counts on disjoint time intervals are independent Poisson random variables with the corresponding means. Reversing those intervals proves that the stationary departure stream is a rate-$\nu$ Poisson process, rather than merely a stream with mean rate $\nu$.

Stationarity matters here. If the queue starts empty, a departure by time $h$ requires both an arrival and a completed service time, so its probability is $O(h^2)$; a rate-$\nu$ Poisson process would have probability $\nu h+O(h^2)$. Thus the output assertion is the equilibrium assertion in the context of the preceding stationary calculation.

Here $x_r$ is the rate per individual active flow, and $n_r{x}_r$ is its route's aggregate rate. On routes with $n_r>0$, a proportionally fair allocation is a feasible positive rate vector satisfyingfor every feasible competitor $\tilde{x}$, where feasibility means ${\sum }_{r:j\in r}{n}_r{\tilde{x}}_r\le C_j$. This is equivalent to maximizing weighted logarithmic utility ${\sum }_{r:n_r>0}{n}_r\log x_r$: its gradient gives the displayed inequality, and the supporting-tangent inequality proves sufficiency. The objective is strictly concave on active rates. Rates assigned to absent flows are irrelevant and may be set to zero.

For the linear flow network, an active local route can increase its rate if its link constraint has slack, improving the objective without affecting any other link. ThusPut $m={\sum }_{i=1}^I{n}_i$ and $N=n_0+m$. If $n_0>0$ and $m>0$, eliminate the active local rates by $x_i=(1-n_0{x}_0)/n_i$. Up to constants, the utility becomesIts derivative is $n_0/x_0-n_{0}m/(1-n_0{x}_0)$, which vanishes exactly atStrict concavity proves this is the optimum, and unused local links impose no stronger constraint because $n_0{x}_0\le 1$. If $m=0<n_0$, maximizing $n_0\log x_0$ gives $x_0=1/n_0$, the same formula. If $n_0=0$, each active local route has $x_i=1/n_i$. These are the proportionally fair allocation on a linear flow network. When $N=0$, there is no service to allocate.

The memoryless property of exponential document sizes makes a flow transmitting at rate $x_r$ complete with intensity ${\mu }_r{x}_r$. Summing over its $n_r$ flows gives the flow-level network model transition ratesand, for nonempty states,All absent-route departure intensities, including those at the empty state, are zero. In particular the local completion rate is the aggregate rate ${\mu }_{i}m/N$, not ${\mu }_i/N$. Total rates are bounded by ${\sum }_r{\nu }_r+{\sum }_r{\mu }_r$, so the continuous-time Markov chain is nonexplosive.

Set ${\rho }_r={\nu }_r/{\mu }_r$ and consider the weightsThe binomial coefficient is the balance function of a flow-level network. For a through-route arrival, its adjacent-weight ratio isThe reverse departure rate is ${\mu }_0(n_0+1)/(N+1)$, so $w(n+e_0)q(n+e_0,n)=w(n){\nu }_0$. For a local-route arrival,and the reverse departure rate is ${\mu }_i(m+1)/(N+1)$, giving the same detailed balance identity. This includes arrivals from the empty state.

To normalize, first fix all local counts and sum over $n_0$. The negative binomial series givesThe remaining geometric series factor, soThis is finite precisely under the stated strict load inequalities ${\rho }_0+{\rho }_i<1$ for each resource. Thus detailed balance proves the stationary law of a linear flow network:With positive arrival rates this is the unique stationary distribution of the irreducible chain. Zero arrival rates are handled by restricting to the accessible class with their counts zero, interpreting $0^0=1$.

Summing this distribution over $n_0$ gives local-count independence in a linear flow network:These are independent geometric distributions. Differentiating a geometric series, or summing its mean directly, givesThis independence concerns the local counts at one stationary time; their evolving processes share the through route and are not independent processes.

The necessary and sufficient condition isEquivalently, $1,u/(2\pi ),v/(2\pi )$ are linearly independent over $\mathbb{Q}$. This is the equidistribution criterion for a torus translation.

For sufficiency, set $w=e^{i(ku+\ell v)}$. For a nonconstant character of the torus the condition gives $w\ne 1$, and the geometric series formula yieldsThe constant torus character has average one. Thus orbit averages converge to integrals against normalized Haar measure for every trigonometric polynomial. Part (i) and a uniform approximation bound extend this to every continuous function $g$: the difference between the orbit average and the integral for $g$ differs from that for an approximating polynomial by at most twice their uniform distance.

For a nonempty half-open rectangle $R$, choose continuous functions $g_{-}\le 1_R\le g_{+}$ with values in $[0,1]$ and $\int (g_{+}-g_{-})dm$ arbitrarily small. One can take distance cutoffs: $g_{-}$ vanishes at the boundary and rises to one inside, while $g_{+}$ is one on the closure and falls to zero outside a thin neighborhood. Their difference is supported in shrinking neighborhoods of the rectangle's finitely many boundary segments, which have Haar measure zero. Squeezing the orbit averages between the two continuous averages gives the normalized area $m(R)=(b-a)(d-c)/(2\pi {)}^2$. Empty rectangles are immediate, and full-circle coordinate intervals can be treated with the constant cutoff in that coordinate.

For necessity, if a nonzero pair is resonant, every orbit point lies in the closed set $H=\{(s,t):e^{i(ks+\ell t)}=1\}$. This is a proper subset because the torus character is nonconstant. Its open complement contains a rectangle of positive area in a coordinate chart, which can be chosen inside the standard coordinate square. That rectangle is never visited, contradicting the required positive limiting frequency. Hence the boxed condition is necessary as well as sufficient.

A power series has radius of convergence $R$ when it converges absolutely for every $|z|<R$ and diverges for every $|z|>R$. Its behaviour on $|z|=R$ is not specified by this definition. When $R=0$, it converges at the centre and diverges at every nonzero point; an infinite radius of convergence means absolute convergence at every finite $z$.

Suppose, towards a contradiction, that $|a_n{z}^n|\le M$ for all $n$ at some $|z|>R$. Choose a complex number $w$ with $R<|w|<|z|$. ThenThe comparison test for series with the convergent geometric series $\sum M{q}^n$ yields absolute convergence at $w$, contradicting the definition of $R$. Thus terms of a power series are unbounded outside its convergence disc:For the given power series, the ratio test gives, when $z\ne 0$,Hence it converges absolutely for $|z|<1$ and its terms fail to tend to zero for $|z|>1$ (the ratios are eventually larger than some number greater than $1$). Therefore $\boxed{R=1}$. In fact, on $|z|=1$ it also converges absolutely by comparison with the convergent p-series $\sum n^{-3}$.

The ratio test says that a series $\sum u_n$ whose terms are eventually nonzero converges absolutely if $\mathrm{limsup}|u_{n+1}/u_n|<1$; in particular, it suffices to have $|u_{n+1}/u_n|\le q<1$ eventually, since then its terms are bounded by a convergent geometric series. The comparison test for series states that, for nonnegative terms, comparison with a larger convergent series gives convergence, and comparison with a smaller divergent series gives divergence. For $u_n=n!/n^n$,By the binomial theorem, $(1+1/n{)}^n\ge 1+n(1/n)=2$. Thus $u_{n+1}/u_n\le 1/2$ for every $n\ge 1$, and induction gives $0<u_n\le 2^{1-n}$. The comparison test for series with $\sum 2^{1-n}=2$ proves

For a spherically symmetric potential, the Poisson equation for Newtonian gravity becomesIntegrate from the center, excluding any additional point mass there. The boundary condition is $r^2{\varphi }^{\prime }(r)\to 0$ as $r\to 0$, soThis derives the enclosed-mass form of the spherical shell theorem from the differential equation. Outside the mass, integrating ${\varphi }^{\prime }=GM/r^2$ gives $\varphi =-GM/r+C$; the conventional zero at infinity sets $C=0$.

For the dyadic shells, multiply their constant mass density by their spherical volumes. The $n$th shell contributesSumming the geometric series yieldsAt radius $r=2^{-N}$, the enclosed shells are exactly those with $n\ge N+1$; a boundary itself contributes no mass. ThusDividing by $r^2=4^{-N}$ givesThese dyadic spherical shells with inverse-radius density scaling have the same field magnitude at each shell boundary, although the field varies inside an individual shell. Finally, outside the planet,with zero potential at infinity; without that convention an arbitrary additive constant remains.

The exponential series at $1$ givesFor every positive integer $q$, multiply by the factorial $q!$ and separate the integer part:Every term in $R_q$ is positive. Each denominator is at least $(q+1{)}^j$, and the inequality is strict when $j\ge 2$. Comparison with the geometric series therefore givesIf $e=p/q$ with integers $p$ and $q\ge 1$, then $q!e=p(q-1)!$ would be an integer. Since $I_q$ is also an integer, $R_q$ would be an integer strictly between zero and one, which is impossible. HenceThis proves the irrationality of e by using its rapidly decreasing series tail.

There is now no separate rate-$\nu$ clockwise transition. The only transitions areBoth preserve $m+n$ modulo $M$. The initial state therefore selects the closed communicating class $m+n\equiv 0(\mathrm{mod}M)$, on which $m\equiv -n(\mathrm{mod}M)$. The population is a birth-death process with birth rate ${\lambda }_{-n}$ and death rate $\mu$.

Define $a_0=1$ and $a_r={\mu }^{-r}{\prod }_{j=0}^{r-1}{\lambda }_{-j}$ for $1\le r<M$. The detailed balance for a birth-death process weights areWriting $n=kM+r$ with $0\le r<M$, the product condition gives $w_n=a_r{\rho }^{kM}$. Summing the resulting geometric series yieldsThis is also the law in part (b) conditioned on $m+n\equiv 0(\mathrm{mod}M)$: the ratio of its weights at consecutive accessible populations is ${\lambda }_{-n}/\mu$. The stability condition is $\mu >\nu$, which uses the product of the periodic birth rates; it does not require $\mu >{\lambda }_m$ for every $m$.

Start from the Erlang loss formula and reverse the order in its denominator:For each fixed $j$, the product tends to $(C/\nu {)}^j$. If $\nu >C$, it is bounded by this same summable geometric series. Extend the summand by zero for $j>NC$ and apply the dominated convergence theorem with counting measure to obtainIf $\nu \le C$, retain the first $K+1$ summands. Their limiting sum is at least $K+1$, so the reciprocal tends to infinity as $K$ can be arbitrarily large. This includes $\nu =C$ without a separate normal approximation. ThereforeHere $C$ and $N$ are positive integers, as appropriate for circuit counts; taking $\lfloor NC\rfloor$ gives the same limit for a positive real capacity scale. This is the proportional scaling limit of the Erlang loss formula.