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For $u_k=a^{2k+1}/(2k+1)>0$, the consecutive ratio isThus the ratio test proves convergence when $0<a<1$ and divergence when $a>1$. At $a=1$, the series is ${\sum }_{k\ge 0}1/(2k+1)$, which diverges by comparison with $\frac{1}{2}{\sum }_{k\ge 0}1/(k+1)$. ThereforeAll terms are nonnegative, so convergence here is also absolute convergence.

The function is even, so each product with a sine is an odd function and all the Fourier coefficients $b_n$ vanish. Its constant coefficient isFor $n\ge 1$, part (a) gives the remaining Fourier coefficients:Consequently the Fourier cosine series isThe periodic extension is continuous, including at $\pm \pi$, and piecewise continuously differentiable. Thus the Fourier series equals $f$ everywhere. Its coefficients are also bounded in magnitude by a constant times $1/n^2$, which gives absolute convergence.

Steady state. In the heat equation, time independence makes the second spatial derivative zero. The two endpoint temperatures therefore determine the linear profileInsulation is along the sides; the endpoint reservoirs can still exchange heat with the rod.

Transient solution. Subtract the steady profile before applying separation of variables: let $U=T-T_{0}x/l$. Then $U_t=\nu U_{xx}$, $U(0,t)=U(l,t)=0$ and $U(x,0)=-T_{0}x/l$ for $0<x<l$. Writing $U=X(x)\Theta (t)$ givesThe homogeneous Dirichlet boundary conditions exclude $\mu \le 0$ and select the eigenfunctions $\sin (n\pi x/l)$ with eigenvalues ${\mu }_n=(n\pi /l{)}^2$, $n\ge 1$. Their orthogonality gives the Fourier sine series coefficientsHere integration by parts uses ${\int }_0^{l}x\sin (n\pi x/l)dx=-l^2(-1{)}^n/(n\pi )$. ThusEvery summand satisfies the heat equation and the homogeneous endpoint conditions, while the steady profile supplies the prescribed endpoint values. The initial Fourier sine series equals $-T_{0}x/l$ in the open interval. At $(l,0)$ the initially cold rod and the suddenly imposed hot endpoint are incompatible; the formula describes the prescribed data for $t>0$ and the initial limit at interior points.

Heat absorption. Integrating in $x$ kills the even modes. With $a=\nu {\pi }^2/l^2$, the total heat relative to the initial state isConsequently the net rate of heat absorption isFor each $t\ge \epsilon >0$, any fixed number of differentiated modes is bounded by a polynomial in $n$ times $e^{-a\epsilon n^2}$. These bounds have finite sums, proving absolute convergence, uniform convergence on such time intervals and the validity of the integration and differentiation. In particular the displayed rate is finite for every $t>0$, positive when $T_0>0$, and tends to zero as equilibrium is approached. The same answer follows from the heat equation balance $H^{\prime }=\gamma \nu [T_x(l,t)-T_x(0,t)]$; this subtracts loss at the cold endpoint from input at the hot endpoint.

Write a nonterminating parity series as $y=x^r{\sum }_{j\ge 0}{b}_j{x}^{2j}$, where $r=0$ or $1$. Its recurrence relation isThe ratio test applied to its nonzero successive terms has limiting absolute ratio $|x{|}^2$. Hence the series converges absolutely for $|x|<1$ and diverges for $|x|>1$, withAt $x=\pm 1$, the limiting ratio is one, so the basic ratio test alone is inconclusive. Retain the next term of the ratio instead:For large $j$ the ratio is positive. Taking logarithms and summing gives $\log |b_j|=-\frac{3}{2}\log j+O(1)$; the $O(j^{-2})$ remainders have a finite sum. Thus $|b_j|\le C{j}^{-3/2}$ eventually. Comparison with the convergent p-series proves absolute convergence at both endpoints as well:The radius of convergence one is the distance from the expansion point to the nearest singular points of a differential equation, $x=\pm 1$. Convergence of function values at those points does not make the solutions analytic there; endpoint derivatives can diverge. The terminating Chebyshev polynomial is an exception with infinite radius of convergence. For example, at $m=0$ the other solution is $\arcsin x$, whose endpoint values are finite while its first derivative diverges.

Consider the counterclockwise unit circle and setThe parametrization $z=e^{i\theta }$ givesThe imaginary part is zero: replacing $\theta$ by $2\pi -\theta$ conjugates the integrand, since $n$ is an integer and $x$ is real. The real part is precisely the defining Bessel integral, so $I=J_n(x)$.

Both exponential series converge absolutely and uniformly on the circle. Their product may therefore be integrated term by term:Uniform absolute convergence can also be checked by summing the absolute coefficient bounds, which gives $e^{|x|}$. The residue is obtained exactly when $n-1+j-k=-1$, or $k=n+j$. The Cauchy coefficient formula consequently yieldsThis proves the Bessel function of the first kind series directly from its integral, including the sign and factorial normalization.

For $x>0$ the geometric ratio $e^{-2x}$ has modulus less than one, andEvery summand is nonnegative. The monotone convergence theorem therefore permits integration of the sum:The integral is finite because its integrand is bounded above by $e^{-x}$, whose integral is one. The displayed paired sums are the even partial sums of the alternating harmonic series; its odd partial sums differ by a term tending to zero. Thus the required equality holds with the usual series ordering. One can additionally evaluate it by $t=e^{-x}$:Grouping into nonnegative pairs supplies the justified interchange; absolute convergence of the ungrouped integrated alternating series is not needed and would be false.

Apply the monotone convergence theorem to $s_N={\sum }_{k=1}^N|a_k|$. Its limit $s$ satisfiesThus $s$ is finite almost everywhere, which proves absolute convergence of ${\sum }_k{a}_k$ almost everywhere. Define its sum $a$ there, and set $a=0$ on the exceptional null set. Then $|a|\le s$, so $a$ is integrable. The partial sums of ${\sum }_k{a}_k$ are also dominated by $s$; dominated convergence givesThe series of integrals converges absolutely, since $|\int a_k|\le \int |a_k|$.

On the square, $|z|\ge N$. The identity $|\sin \pi (x+\alpha +iy){|}^2={\sin }^2\pi (x+\alpha )+{\sinh }^2\pi y$ bounds the sine away from zero on the vertical sides because $x=\pm N$; on horizontal sides it grows exponentially. Therefore the whole contour integral is $O(N^{-1})$ or smaller and vanishes. The residue theorem then giveswhich proves $\boxed{\sum _{n\in \mathbb{Z}}(-1{)}^n/(n-\alpha {)}^2={\pi }^2\csc \pi \alpha \cot \pi \alpha }$. Absolute convergence justifies the limiting residue sum.

For the Fourier sine series write $u_0(x)={\sum }_{n\ge 1}{b}_n\sin (n\pi x/L)$. Orthogonality of trigonometric functions givesIn the second half of the integral put $z=L-x$; then $u_0(L-z)=\rho z$ and $\sin [n\pi (L-z)/L]=(-1{)}^{n+1}\sin (n\pi z/L)$. HenceEvery even coefficient is zero. For odd $n$, part (a) or the supplied integral yieldsPutting $n=2m+1$ therefore givesThis is the Fourier sine series of a symmetric tent. The absolute values of its terms are uniformly bounded by a summable sequence, giving absolute convergence and uniform convergence; the piecewise smooth odd extension and the Dirichlet convergence theorem for Fourier series identify its value with the continuous $u_0$, including the zero endpoints and the peak.

The printed lower limit $m=1$ is an error. The original PDF visibly has this lower limit. It omits the nonzero fundamental coefficient $F_0=4L\rho /{\pi }^2$. In particular, the printed series equals $u_0(x)-F_0\sin (\pi x/L)$, rather than $u_0(x)$. At $x=L/2$ it gives $L\rho /2-4L\rho /{\pi }^2$ instead of $L\rho /2$. The corrected lower limit is displayed explicitly above.

The Poisson summation formula equates sample sums of a function and its Fourier transform. Exchanging broad and narrow sums can save work. Use the conventionFor every Schwartz function $f\in \mathcal{S}(\mathbb{R})$,Both series have absolute convergence. A sufficient hypothesis is that $f$ is smooth and ${\sup }_x(1+|x|{)}^M|f^{(r)}(x)|<\infty$ for all nonnegative integers $M,r$; its Fourier transform is then also a Schwartz function.^[ref] This assumption controls both decay and regularity; mere membership of an L1 space does not control values at individual sampling points, as periodization of an integrable function illustrates.

Here $\xi$ counts cycles per unit length. In the angular convention used elsewhere in Fourier analysis, $\tilde{f}(\omega )=\int f(x)e^{-i\omega x}dx$, so $\hat{f}(\xi )=\tilde{f}(2\pi \xi )$ and the right-hand side becomes ${\sum }_k\tilde{f}(2\pi k)$. Keeping the Fourier transform convention explicit prevents stray $2\pi$ factors. Changing the sample spacing leads to the Poisson summation formula for a Euclidean lattice.

Let $\Lambda =B{\mathbb{Z}}^d$ be a full-rank Euclidean lattice, with covolume $m(\Lambda )=|\det B|$ and dual lattice ${\Lambda }^{\vee }=B^{-\mathsf{T}}{\mathbb{Z}}^d$. For a Schwartz function $f\in \mathcal{S}({\mathbb{R}}^d)$ use the Fourier transform conventionThen the Poisson summation formula becomesBoth series have absolute convergence. To see the normalization, apply integer-lattice Poisson summation to $g(u)=f(Bu)$. The change of variables formula gives $\hat{g}(k)=|\det B{|}^{-1}\hat{f}(B^{-\mathsf{T}}k)$, yielding both the reciprocal covolume factor and the dual lattice argument. The factor is the physical lattice-point density. In particular, for spacing $h>0$,For the Gaussian function $f(x)=e^{-\pi a\Vert x{\Vert }^2}$ with $a>0$, the product of the one-dimensional Gaussian Fourier transforms givesHere $\Vert \cdot \Vert$ is the Euclidean norm. As a normalization check, take $\Lambda =\mathrm{diag}(2,3){\mathbb{Z}}^2$. Its covolume is $6$, and its dual lattice is $\mathrm{diag}(1/2,1/3){\mathbb{Z}}^2$, soThis also follows by multiplying two Gaussian Poisson summation identities with parameters $4a$ and $9a$. If one instead uses an angular Fourier transform kernel $e^{-i⟨x,\omega ⟩}$, the reciprocal frequencies are $2\pi {\Lambda }^{\vee }$, while the prefactor remains $m(\Lambda {)}^{-1}$.

The radius of convergence of ${\sum }_{n\ge 0}{a}_n(z-c{)}^n$ is the unique $R\in [0,\infty ]$ separating interior absolute convergence from exterior divergence. It converges absolutely when $|z-c|<R$ and diverges when $|z-c|>R$. At the centre it always has the finite value $a_0$, since all positive powers vanish. The boundary convergence of a power series is a separate question.

For $0<R<\infty$, the interior is a disk in the complex plane and an open interval $(c-R,c+R)$ on the real line. A power series converges uniformly on every smaller closed disk $|z-c|\le r<R$. This gives stable polynomial approximation over a whole compact region, rather than a separate limiting calculation at each point. See Jiří Lebl, Basic Analysis, §11.3.2: convergence of power series.

Four examples reveal what $R$ measures:

The Cauchy-Hadamard theorem computes $R$ directly from the coefficients. The attractive feature is that one growth statistic decides convergence throughout the interior and exterior, including sparse series with many zero coefficients. Finite changes to the coefficients do not alter the radius, though they can change the sum.

Let $y^{\prime \prime }+{\omega }^{2}y=F(t)$ with $\omega >0$, zero initial displacement and velocity, and real-valued forcing with ${\int }_0^{\infty }|F(t)|dt<\infty$. The causal Green function gives $y(t)={\omega }^{-1}{\int }_0^{t}F(s)\sin [\omega (t-s)]ds$. Expanding the sine difference, the two coefficients tend to finite limits by absolute convergence. Thus $y(t)$ approaches a sinusoid whose amplitude is the displayed modulus of the one-sided Fourier transform, divided by $\omega$. The displacement error is at most ${\omega }^{-1}{\int }_t^{\infty }|F(s)|ds$, which tends to zero. This describes the surviving free oscillation after a transient force; an arbitrary nonzero initial oscillation must be added before taking the final amplitude.

For $a>0$, let $g(x)=f(ax)$. Whenever the Mellin transform of $f$ is absolutely convergent,The Mellin fundamental strip is unchanged: the corresponding integrals of absolute values differ by the finite positive factor $a^{-\sigma }$. The change of variables $y=ax$ gives $x^{s-1}dx=a^{-s}{y}^{s-1}dy$, proving the rule. For $s=\sigma +i\tau$, the transform's absolute value changes by $a^{-\sigma }$, while the factor $e^{-i\tau \log a}$ changes its phase. On the line $\sigma =0$, wherever absolute convergence holds, the transform's absolute value is preserved.

For the Mellin transform of an interval indicator, replacing $1_{[1,4]}(x)$ by $1_{[1,4]}(2x)=1_{[1/2,2]}(x)$ givesAt the removable singularity $s=0$, both sides have value $\log 4$. Scaling leaves the width measured by the natural logarithm unchanged. Likewise, the exponential function $e^{-ax}$ has Mellin transform $a^{-s}\Gamma (s)$ on $\mathrm{Re}s>0$.

The two identities areTo prove them, expand $e^{\pm x\log x}$ in its exponential series. Since $|x\log x|\le 1/e$ on $(0,1)$, the Weierstrass M-test justifies termwise integration. Substituting $x=e^{-t}$ gives ${\int }_0^1{x}^k(\log x{)}^{k}dx=(-1{)}^{k}k!/(k+1{)}^{k+1}$ through the gamma function. Inserting these moments into the series and shifting $n=k+1$ proves both formulas. Each resulting numerical series has absolute convergence.

Write $a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n$, $a_0=1$. The binomial series for $\sqrt{1-u}$ has positive numbers $b_n=a_{n-1}-a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/[4^n(2n-1)]$ as the magnitudes of its nonconstant coefficients. The product $a_n={\prod }_{j=1}^n[1-1/(2j)]$ tends to zero, since $\ln a_n\le -\frac{1}{2}{\sum }_{j=1}^{n}1/j\to -\infty$. Thus $\sum b_n=1$ by telescoping. The Weierstrass M-test gives uniform convergence and absolute convergence on $[-1,1]$. Continuity extends the interior identity to both endpoints. Its radius of convergence is one; outside this interval its terms do not tend to zero.

Inside both convergence discs, the triangle inequality gives absolute convergence of the sum. Suppose $R<S$ and the sum had radius $T>R$. At any modulus between $R$ and $\min (T,S)$, subtracting the absolutely convergent second series from the absolutely convergent sum would give absolute convergence of the first series outside its radius of convergence, a contradiction. Interchanging the two series covers $S<R$.

Let a power series ${\sum }_{n\ge 0}{a}_n{z}^n$ have finite radius of convergence $R$. If its terms were bounded in complex modulus at a point $z$ with $|z|>R$, say $|a_n{z}^n|\le M$, choose $w$ with $R<|w|<|z|$. ThenThe comparison test for series with a convergent geometric series would give absolute convergence at $w$, contradicting its position outside the convergence disc. Hence the terms are unbounded at every such $z$. The argument also works when $R=0$ and gives a stronger obstruction than the usual term test for divergence.