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Expand the finite Euler product into geometric series. By unique prime factorization, it is the sum of $1/n$ over positive integers whose prime factors do not exceed $x$, and hence contains every term up to $m=\lfloor x\rfloor$. The harmonic sum satisfiesThe last inequality uses $m+1>x$, including when $x$ is an integer. For each prime number $p$,The telescoping series over all integers $n\ge 2$ bounds the sum of these remainders by $\frac{1}{2}{\sum }_{n\ge 2}[n(n-1){]}^{-1}=1/2$. Taking logarithms of the product bound therefore givesThis explicit constant suffices; no asymptotic theorem about the distribution of prime numbers is needed.

The radius of convergence of the power series ${\sum }_{n\ge 0}{a}_n{z}^n$ isIf $|z|<R$, choose $w$ with $|z|<|w|<R$. Since $a_n{w}^n\to 0$, those terms are bounded, say $|a_n{w}^n|\le M$. Thereforeand comparison with a geometric series proves absolute convergence. If $|z|>R$, convergence would imply absolute convergence at every smaller modulus by the same argument, contradicting the definition of $R$.

If $|a_n|\le |b_n|$, convergence of $\sum b_n{w}^n$ implies absolute convergence of $\sum a_n{z}^n$ whenever $|z|<|w|$. Letting $|w|$ approach the radius of the $b$-series shows that $R_a\ge R_b$.

Choose $x\in X$ and iterate $x_{n+1}=f(x_n)$. The contraction mapping estimatemakes $(x_n)$ Cauchy by a geometric series. Its limit $x_0$ satisfies $f(x_0)=x_0$ because $f$ is Lipschitz. Two fixed points would satisfy $d(x_0,y_0)\le \lambda d(x_0,y_0)$, so they coincide. This proves the Banach fixed-point theorem.

Now suppose the final assertion failed. There would be $m_k,n_k\to \infty$ with $f(a_{m_k})=a_{n_k}$. Since $a_n\to a$ and $f$ is continuous, the two sides tend respectively to $f(a)$ and $a$, so $f(a)=a$. Uniqueness would give $a=x_0$, a contradiction. Hence the required $\ell$ and $m$ exist.

For an increasing sequence $(n_k)$ of nonnegative integers, full-density sequence means that its range $S$ has natural density one:For a sequence $(z_n)$ in a metric space, convergence in density of a sequence to $z$ means that, for every $\epsilon >0$,In particular, a bounded nonnegative scalar sequence converging to zero in density has averages tending to zero: the average is at most $\epsilon$ plus its bound times the proportion of indices above $\epsilon$.

Here is a Hilbert space form of the Van der Corput lemma, including the density version. Suppose $\Vert u_n\Vert \le M$, and putThenSince $0\le c_h\le M^2$, it suffices for $(c_h)$ to converge to zero in density. A frequently used special case is that the correlation average tends to zero for every fixed $h\ge 1$.

To prove the lemma, extend the finite list $u_0,\dots ,u_{N-1}$ by zero outside this range, and write $S_N={\sum }_{n=0}^{N-1}{u}_n$. For $1\le H\le N$,The Cauchy-Schwarz inequality for this sum, followed by expansion of the squared Hilbert space norm, givesThe coefficient $H-h$ counts the pairs of shifts separated by $h$. Dividing by $H^2{N}^2$, bounding real parts by absolute values, and first taking the limit superior in $N$ for fixed $H$ yieldsThe right side tends to zero as $H\to \infty$ under the stated hypothesis. This proves both the lemma and its density consequence.

For the polynomial application, fix a nonzero integer $m$ and let $u_n=e^{2\pi imP(n)}\in \mathbb{C}$. For each fixed $h\ge 1$, the correlation phase is linear:Thus, with $q_h=e^{2\pi i(2mah)}$,The coefficient $2mah$ is irrational, so $q_h\ne 1$. This geometric series is bounded in modulus by $2/(N|1-q_h|)$ and tends to zero. All $c_h$ vanish, so the Van der Corput lemma gives

The Weyl criterion now shows that the fractional parts form an equidistributed sequence. For completeness, its sufficiency here follows directly: the displayed limits give the correct average for every trigonometric polynomial, including the constant term. The Stone-Weierstrass theorem extends this to every continuous function by uniform approximation. Approximating an interval's indicator function above and below by continuous functions gives its length as the limiting frequency. HenceThis proves the equidistribution of a quadratic polynomial with irrational leading coefficient using the required correlation argument.

For the K-theory transfer of a finite covering, let $V\to Y$ be a vector bundle and defineOn an evenly covered neighborhood the $n$ sheets identify this with a direct sum of $n$ local vector bundles. On overlaps the identifications permute the sheets and apply the transition maps of $V$, so these local descriptions glue to a vector bundle on $X$. This operation respects direct sums, hence extends to a homomorphism of Grothendieck groupsIt obeys the projection formula for the K-theory transfer:For actual vector bundles, this follows fiberwise by distributing the tensor product of vector bundles over the direct sum; it then extends to their Grothendieck groups.

Let $P=p_{!}(1_Y)$ be the permutation vector bundle of the covering. It has rank $n$ on every connected component, but need not be a trivial vector bundle; in particular one cannot replace $p_{!}{p}^{\ast }$ by multiplication by $n$ integrally. Instead writeThe componentwise nilpotence result from the preceding solution gives ${\eta }^N=0$ for some $N$. In the localization of a ring $K^0(X)\otimes \mathbb{Z}[1/n]$, the finite geometric seriesis an inverse. The projection formula for the K-theory transfer givesThus $p^{\ast }(a)=0$ implies $a=0$ after inverting $n$. More explicitly, $(n+\eta {)}^{-1}{p}_{!}$ is a left inverse, proving

Write $y={\psi }_s$ and retain the Landweber relaxation parameter $\tau$. The printed $\Vert A\Vert <2$ allows the fixed choice $\tau =1/2$, since then $\tau \Vert A{\Vert }^2<2$. It does not justify a unit step: the scalar linear operator $A=3/2$ has norm below $2$, but unit-step error is multiplied by $1-|A{|}^2=-5/4$ and diverges. We use $0<\tau <2/\Vert A{\Vert }^2$ throughout.

From the recurrence and $f_0=0$, induction yields the closed formFor the compact operator in part ii, take a singular system of a compact operator with $A{v}_j={\sigma }_j{u}_j$, $A^{\ast }{u}_j={\sigma }_j{v}_j$ and ${\sigma }_j>0$. Thus ${\sigma }_j^2$ are the eigenvalues of $A^{\ast }A$; this fixes the notation implicit in the printed hint. In the singular component $j$, the recurrence isSumming the geometric series gives the Landweber spectral filterThere is no contribution from data in $\ker A^{\ast }$ or from $\ker A$ in the reconstruction. Starting at zero is what selects the minimum-norm least-squares solution.

For fixed ${\sigma }_j>0$, $|1-\tau {\sigma }_j^2|<1$, so the numerator tends to $1$. The assumption $y\in D(A^{†})$ is the Picard criterionwith an arbitrary additional component in $\ker A^{\ast }$. The squared reconstruction error isEach term tends to zero and is bounded by a summable term from the Picard criterion. The dominated convergence theorem therefore proves $R_{n}y\to A^{†}y$, where $A^{†}$ is the Moore–Penrose inverse of an operator.

For each finite $n$, the regularization of an inverse problem is stable. Set $t=\tau {\sigma }^2\in (0,2)$. The geometric series and $|1-t|\le 1$ implyConsequently the Landweber noise amplification bound in this general step-size convention is $\Vert R_n\Vert \le \sqrt{2\tau n}$. If additionally $\tau \Vert A{\Vert }^2\le 1$, the sharper numerator bound $\min (nt,1)$ gives $\Vert R_n\Vert \le \sqrt{\tau n}$.

To see the regularization parameter directly, put $\alpha =1/(\tau n)$. Modes with ${\sigma }^2\ll \alpha$ have numerator approximately $n\tau {\sigma }^2$, so their inverse coefficient is approximately $\sigma /\alpha$ rather than $1/\sigma$. Each fixed nonzero mode is eventually restored as $\alpha \downarrow 0$. ThusThe finite iterates suppress unstable small singular values; infinitely many iterations remove this suppression. For noisy data $\Vert y^{\delta }-y\Vert \le \delta$, the noise-bias decomposition for linear regularization givesChoosing $n(\delta )\to \infty$ and ${\delta }^{2}n(\delta )\to 0$ proves noisy-data convergence. The reciprocal iteration index is a regularization parameter, and early stopping controls noise amplification.

For each one-particle state $i$, a boson occupation number can be any $n_i\ge 0$. Its contribution to the grand canonical partition function is the geometric seriesThe noninteracting modes factorize, soConvergence requires $\mu <{\min }_i{ϵ}_i$. Since the lowest energy is zero, the limiting maximum chemical potential is

LetThe set $S$ contains zero. If $r\in S$ and $0\le s<r$, then $s\in S$ by comparison. Moreover, if the original power series diverges at $x_0$, no $r>|x_0|$ can lie in $S$, because absolute convergence at $r$ would imply absolute convergence at $x_0$. Thus $R=\sup S$ is finite and nonnegative.

If $|x|<R$, choose $r\in S$ with $|x|<r$; comparison gives absolute convergence at $x$. If $|x|>R$ and the series converged at $x$, then its terms $a_n{x}^n$ would be bounded. For any $y$ with $R<|y|<|x|$,so the geometric series comparison would put $|y|$ in $S$, contradicting the definition of $R$. This proves the radius of convergence property.

For ${\sum }_{n\ge 1}{x}^n/3^n$, the ratio is $x/3$, so the geometric series criterion gives

A Laurent series about $a$ is an expansionthat converges to $f$ on an annulus $r<|z-a|<R$.

The partial fraction decomposition isFor $0<|z|<1$, expand both denominators by the geometric series:This is actually a Taylor series at zero, since the apparent puncture at zero contains no singularity.

For $1<|z|<2$, the $z+2$ term still expands in nonnegative powers, while the quadratic term must be expanded in negative powers:Hence

On each left singular vector $u_j$, use $A^{\ast }{u}_j={\sigma }_j{v}_j$ and $(I-A^{\ast }A)v_j=(1-{\sigma }_j^2)v_j$. The finite geometric series givesThus with $\alpha =1/n$, the dimensionless Landweber spectral filter isThe full inverse-coefficient convention instead uses $[1-(1-{\sigma }^2{)}^{1/\alpha }]/\sigma$. With general step $\tau$, replace ${\sigma }^2$ inside the power by $\tau {\sigma }^2$.

In the normalized problem, $0\le f_n(\sigma )\le \min (n{\sigma }^2,1)$, so the inverse coefficient is at most $\min (n\sigma ,1/\sigma )\le \sqrt{n}$. This verifies directly why finite iteration is stable and why admitting progressively smaller singular values eventually amplifies noise.

Write $B$ for the open unit ball. Because $T$ is a linear map, the hypothesis scales toFix $y\in B$. Choose numbers $r$ and $\theta$ such thatthen $r/(1-\theta )<1$.

Set $z_0=y$. Inductively, if $\Vert z_{j-1}\Vert <r{\theta }^{j-1}$, the scaled density statement lets us chooseand we put $z_j=z_{j-1}-T{x}_j$. The geometric series givesBecause $X$ is a Banach space, $x={\sum }_{j\ge 1}{x}_j$ exists and lies in $B$. Since $T$ is a bounded linear operator, it is continuous, whileTherefore $Tx=y$. As $y\in B$ was arbitrary,This is the dense open-unit-ball image criterion.

Letand choose $0<\eta <\delta$. Write $C_{\eta }$ for the positively oriented boundary of the square with $|x|,|y|\le 1+\eta$. The Cauchy integral formula gives, for every $z\in S$,Because $C_{\eta }$ and $S$ are compact and separated by a positive distance, Riemann sums for this contour integral converge uniformly in $z\in S$. Thus, for any $\epsilon >0$, there are points ${\zeta }_j\in C_{\eta }$ and constants $c_j$ such that

It remains to approximate each resolvent by polynomials on $S$. Let $\Lambda$ be the set of $\lambda \notin S$ for which $(z-\lambda {)}^{-1}$ is polynomially uniformly approximable on $S$. If $|\lambda |>{\sup }_{z\in S}|z|$, the geometric seriesshows that $\lambda \in \Lambda$. Moreover, the resolvent propagation across a complementary component argument shows that $\Lambda$ is both open and closed relative to $\mathbb{C}\setminus S$. The complement of a square is connected, and $\Lambda$ is nonempty, so $\Lambda =\mathbb{C}\setminus S$.

In particular, every ${\zeta }_j$ lies in $\Lambda$. Approximate the finitely many resolvents accurately enough that their weighted sum differs by less than $\epsilon /2$. Combining this with the Riemann-sum estimate produces a polynomial $p$ satisfyingThis proves the claimed elementary polynomial approximation on a square.

For $x\ge 2$, expand the finite Euler product as a product of geometric series:Every integer $1\le n\le x$ has all its prime factors at most $x$, soThe harmonic series diverges, and therefore

Suppose instead that ${\sum }_{p}1/p$ converged. For $0<t\le 1/2$,Since ${\sum }_{p}1/p^2$ converges, this would makeconverge to a finite limit, contradicting the divergence of $P(x)$. Hence the Euler proof that the sum of reciprocals of primes diverges gives

The partial fraction decomposition isFor $0<|z|<2$, expand the second term as a geometric series:Hence

The polynomial Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $p$ such that ${\sup }_{z\in K}|p(z)-f(z)|<\epsilon$.

For an explicit approximation on the left semicircle $K=\{z:|z|=1,\mathrm{Re}z\le 0\}$, defineThe two power-series expansionsproduce this formula. The first is a geometric series whose ratio has modulus at most $4/\sqrt{17}<1$ on $K$. The second converges uniformly on the whole unit circle; the elementary bound $\left(\genfrac{}{}{0ex}{}{j+k}{k}\right)\le 2^{j+k}$ shows that truncating it at $k=n^2$ gives a total error tending to zero even after summing over $0\le j\le n$. Consequently $p_n\to 1/z$ uniformly on $K$. This is an explicit polynomial approximation of the reciprocal on the left semicircle.

No such sequence exists on the punctured unit circle $S=\{z:|z|=1,z\ne 1\}$. If polynomials $p_n$ converged uniformly there, they would be Uniformly Cauchy. Continuity givesso they would converge uniformly on the entire unit circle. The uniform limit theorem would force the value at the missing point to be $1$, and hence the limit would be $1/z$ on the full circle. But the Cauchy integral theorem gives $\oint p_n(z)dz=0$ for every $n$, while uniform convergence and contour integration would implya contradiction. This is the polynomial approximation obstruction on a punctured circle.

Put $q=e^{-\lambda t}$ and propose $P_n=q(1-q{)}^{n-1}$. The initial condition follows from $q(0)=1$. Since $q^{\prime }=-\lambda q$,On the other hand,and expanding the bracket gives the same expression. The formula also sums to one by the geometric series, so it is the required probability mass function.

If $\alpha \ne \beta$, the operator $f-\beta I$ is invertible on $W_{\alpha ,n}$. Indeed, there it equalswhose inverse is the finite geometric series

Now use induction on $d$. If ${\sum }_{i=1}^d{v}_i=0$, apply $(f-{\alpha }_{d}I{)}^n$. The $v_d$ term vanishes, while every transformed vectoris nonzero by the invertibility just proved and still lies in $W_{{\alpha }_i,n}$. The induction hypothesis rules out the resulting relation. Allowing scalar coefficients, and omitting zero terms, gives the same argument. Hence the generalized eigenspaces for distinct eigenvalues form a direct sum, so $v_1,\dots ,v_d$ are linearly independent.

Fix $a\notin [\gamma ]$ and letFor $|z-a|<d$ and $\lambda \in [\gamma ]$, the geometric series givesThe series converges uniformly on every smaller closed disc, so it may be integrated term by term along the curve. Hencewhich is a power series about $a$.

If $f$ is holomorphic on a neighbourhood of the closed disc $\overline{D}(a,r)$, the Cauchy integral formula saysDifferentiation under the integral is valid uniformly on smaller discs and gives the Cauchy derivative formulaThis proves inductively that every holomorphic function has complex derivatives of every order. In particular,

Now suppose $f_n\to f$ locally uniformly on $U$. Given a compact set $K\subset U$, choose finitely many closed discs whose slightly larger concentric discs remain in $U$ and whose interiors cover $K$. Applying the derivative formula to $f_n-f$ on the larger boundary circles gives a uniform Cauchy estimatewhere $L\subset U$ is the compact union of those circles. The right-hand side tends to zero, proving that

Finally, choose open neighbourhoods $U_j$ of the closed discs $D_j$ so small thatlies in the given neighbourhood on which $f$ is holomorphic. Inside that neighbourhood choose a positively oriented piecewise smooth contour $\Gamma$ surrounding $D_1\cap D_2$. It may be chosen as the boundary of a slightly enlarged lens and split into arcsso that ${\Gamma }_1$ stays away from $D_1$ and ${\Gamma }_2$ stays away from $D_2$. DefineandBecause each arc avoids the corresponding disc, these formulas define holomorphic functions on possibly smaller neighbourhoods $U_1,U_2$. On their overlap, the Cauchy integral formula for the full contour gives

Repeated substitution in Landweber iteration from $u_0=0$ givesOn the $i$th singular vector, $A^{\ast }A$ has eigenvalue ${\sigma }_i^2$ and $A^{\ast }{y}_i={\sigma }_i{x}_i$. The finite geometric series therefore givesThis is a spectral regularization method with filter

For bosons the occupation number is $n_{\mathbf{p}}=0,1,2,\dots$. Summing the geometric series for the single-mode grand canonical ensemble gives the Bose-Einstein distribution