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Knowing the radius of convergence does not settle convergence on its boundary. The three power seriesall have radius one, but behave differently on $|z|=1$.

For the geometric series, boundary terms all have complex modulus one, violating the term test for divergence. For the second series, absolute values give the divergent harmonic series at every boundary point. At $z=1$ the series itself is harmonic and diverges. At any other point on the unit circle, its geometric partial sums obeyThe Dirichlet test with the decreasing sequence $1/n\to 0$ therefore gives conditional convergence. In particular, $z=-1$ gives the negative of the alternating harmonic series, with sum $-\ln 2$. For the third series, the convergent p-series ${\sum }_{n\ge 1}1/n^2$ bounds the absolute values at every boundary point, giving absolute convergence by the comparison test for series.

At $z=-1$, the rational function $1/(1-z)$ is well defined and equals $1/2$, yet its geometric series diverges there. A formula for a function and a particular series representation can have different domains. Likewise, applying termwise differentiation of a power series preserves the radius but need not preserve the boundary convergence.

Define $h(x)=|x|\log (e\pi /|x|)$ on $[-\pi ,\pi ]$, set $h(0)=0$, and extend periodically. Concavity and monotonicity on $[0,\pi ]$ give $\omega (h,\delta )=\delta \log (e\pi /\delta )$ for $0<\delta \le \pi$. Its cosine Fourier coefficients areOne integration by parts gives the formula; the remaining integral is $\log j+O(1)$ by the Dirichlet test. All these coefficients are nonpositive. The de la Vallée Poussin sum $V_n=2{\sigma }_{2n}-{\sigma }_n$ reproduces degree-at-most-$n$ trigonometric polynomials and has operator norm at most three. Hence $L_n(f)=f(0)-V_{n}f(0)$ has norm at most four and annihilates those polynomials. Its weights on coefficients are zero up to $n$, nonnegative thereafter, and one from $2n$ onwards, so $|L_n(h)|\ge {\sum }_{j\ge 2n}|a_j|\gtrsim \log n/n$. Thus $E_n(h)\ge |L_n(h)|/4\gtrsim \log n/n$; the first Jackson theorem for periodic approximation proves the matching upper bound. This cusp and the critical Weierstrass modulus have the same first-modulus order but different best-approximation rates.

All three examples below have radius of convergence one, because the $n$th roots of their coefficient moduli tend to one.

For convergence everywhere on the boundary, take ${\sum }_{n=1}^{\infty }{z}^n/n^2$. On $|z|=1$ it converges absolutely: for $n\ge 2$, $1/n^2\le 1/[n(n-1)]=1/(n-1)-1/n$, whose sum telescopes to a finite value.

For divergence everywhere on the boundary, take ${\sum }_{n=0}^{\infty }{z}^n$. If $|z|=1$, every term has modulus one, so the term test for divergence applies.

For mixed boundary behavior, take ${\sum }_{n=1}^{\infty }{z}^n/n$. At $z=1$ it is the divergent harmonic series; each successive dyadic block contributes at least $1/2$. At every other point $|z|=1$, summation by parts proves convergence. Indeed put $B_k={\sum }_{j=m}^k{z}^j$, so the finite geometric series gives $|B_k|\le 2/|1-z|=:C_z$. ThenThus the partial sums form a Cauchy sequence and converge. This derives the relevant Dirichlet test rather than only naming it.

Jordan's lemma states that, for $b>0$, if $g(z)\to 0$ uniformly on large upper semicircles, then $\int g(z)e^{ibz}dz\to 0$ on those arcs. Indeed its absolute value is at most $R\max |g|{\int }_0^{\pi }{e}^{-bR\sin \theta }d\theta \le (\pi /b)\max |g|$. For negative $b$, use lower semicircles. The corresponding exponential-decay estimates can also be used on rectangular closures.

First take $a>0$ and $0<\epsilon <a$, and defineUse finite strip rectangles with vertical sides at $\pm R$, $R=N+1/2$, then let $N\to \infty$. Their orientation is positive. At a nonzero integer, the residue isThe origin is removable when $a\ne 0$ and contributes zero. The short vertical-side integrals tend to zero since $|\sin \pi (\pm R+iy)|=\cosh (\pi y)$ and $F=O(R^{-1})$ there. The residue theorem therefore gives the first evaluation,The sum is understood as the limit of symmetric partial sums. It converges for $|x|<\pi$ by the Dirichlet test: the oscillatory partial sums are bounded, and $n/(a^2+n^2)$ decreases eventually to zero.

For the second evaluation, close the top line outward into the upper half-plane and the bottom line outward into the lower half-plane. Because of the strip orientations, both outward closures are clockwise. The upper region contains only the pole $ia$, and the lower one only $-ia$. Their residues areTo justify discarding the added boundaries, use rectangles with far vertical sides at $\pm (N+1/2)$ and outer horizontal sides at imaginary height $\pm R$. The numerator grows at most like $e^{|x||\mathrm{Im}z|}$, while $1/\sin \pi z$ decays like $e^{-\pi |\mathrm{Im}z|}$. Thus the horizontal integral tends exponentially to zero, and the vertical integrals are bounded by a constant times $R^{-1}{\int }_{\epsilon }^R{e}^{-(\pi -|x|)y}dy\to 0$. The strict condition $|x|<\pi$ is essential. The clockwise residue calculation givesEquating the two evaluations proves the alternating sine series with a quadratic denominator:Both sides are even functions of $a$, so this proves the result for every nonzero real $a$. At $a=0$ the displayed zero-index summand and right-hand ratio are literally undefined; under their natural continuous extension, set that summand to zero and take the right-hand limit $-x$. This limit may be passed through the nonzero-index sum, since its difference from the $a=0$ sum is bounded by $2a^2{\sum }_{n\ge 1}{n}^{-3}$. Thus the intended limiting identity at zero is $2{\sum }_{n\ge 1}(-1{)}^n\sin (nx)/n=-x$. The separate contour evaluation above assumes distinct off-axis poles and should not be applied unchanged after they coalesce at the origin.

Take $a=5$ and $x=\pi /2$. Since $5^k\equiv 1(\mathrm{mod}4)$, every cosine term vanishes at $x$, andSet $h=1/n$ and $m=\lfloor {\log }_{5}n\rfloor$. For $0\le k\le m$, the sine argument lies in $[0,1]$, where $\sin u\ge (\sin 1)u$. The initial part therefore has a common sign and magnitude at least $(m+1)(\sin 1)/n$. The remaining tail has absolute value at mostThe reverse triangle inequality now givesfor sufficiently large $n$, because $m+1>\ln n/\ln 5$. This establishes the sharp logarithmic lower boundThe constant can be reduced to include the finitely many small indices: for $n\ge 5$ the displayed lower bound is positive; for $n=2,3$ the first sine exceeds the tail bound $1/4$, and for $n=4$ the first two sines contribute $\sin (1/4)+\sin (5/4)/5$, exceeding the remaining tail bound $1/20$. At $n=1$ the logarithmic right side is zero.

Here $E_n(g)\asymp 1/n$ but $g$ is not Lipschitz continuous, so a first modulus of continuity bound $O(1/n)$ is sufficient by the first Jackson theorem for periodic approximation but is not necessary. Nor is the weaker necessary bound $O(\ln n/n)$ sufficient: an explicit comparison is the logarithmic cusp with slow trigonometric approximationextended periodically. On $[0,\pi ]$ it is increasing and concave, so its largest increment over distance $\delta$ is from zero, giving $\omega (h,\delta )=\delta \ln (e\pi /\delta )$ for $0<\delta \le \pi$.

For a concrete approximation lower bound, its cosine Fourier coefficients satisfyIndeed, integration by parts once replaces $h$ by $h^{\prime }(x)=\ln (\pi /x)$; integrating that against sine with primitive $(1-\cos jx)/j$ yields the stated positive integral. Its logarithmic growth follows because ${\int }_1^A\cos u/udu$ stays bounded by the Dirichlet test. All $a_j$ are negative and absolutely summable.

Let $V_n=2{\sigma }_{2n}-{\sigma }_n$, a de la Vallée Poussin sum. Its multiplier is one for $j\le n$, $2-j/n$ for $n<j<2n$, and zero for $j\ge 2n$; the Fejér summation is a uniform-norm contraction property gives $\Vert V_n\Vert \le 3$. Thus the functional $L_n(f)=f(0)-V_{n}f(0)$ has norm at most four and vanishes on every degree-at-most-$n$ trigonometric polynomial. Because the remaining multiplier weights are nonnegative and all $a_j<0$,The first Jackson theorem for periodic approximation gives the corresponding upper bound. The same first-modulus order $\delta \ln (1/\delta )$ occurs with both $E_n\asymp 1/n$ and $E_n\asymp \ln n/n$. Therefore first-modulus rates alone do not characterize the endpoint approximation class, unlike the matched direct and inverse rates for $0<\alpha <1$.

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.

The endpoint agreement $f(1)=f(0)=0$ makes the boundary terms vanish when integrating the Fourier coefficients of $f^{\prime }$ by parts. If $a_n^{\prime },b_n^{\prime }$ denote these coefficients, thenThese are exactly the coefficients obtained by differentiating the preceding Fourier series:On every compact subinterval of $(0,1)$ the differentiated series converges uniformly by the Dirichlet test, since partial sums of the sine factors are uniformly bounded there. The Termwise differentiation of a Fourier series is therefore valid away from integers and gives $1-2x$ on $(0,1)$. At an integer the periodic original has a corner, so the differentiated series is not asserting the existence of an ordinary derivative there.

Oddness gives zero cosine coefficients. Integration by parts gives the sine Fourier coefficientsThe Fourier series of the piecewise smooth periodic extension converges to its value at interior continuity points. Termwise integration from 0 to $x$ is valid: on the interval between these points, which stays inside $(-\pi ,\pi )$, the partial sums of $\sum (-1{)}^{n+1}\sin (nt)$ are uniformly bounded by the geometric-sum formula, since $|1+e^{it}|$ is bounded away from zero. The Dirichlet test therefore gives uniform convergence there. Integrating givesThe integrated series is absolutely and uniformly convergent. ConsequentlyIndependently, the constant Fourier coefficient is $a_0={\pi }^{-1}{\int }_{-\pi }^{\pi }{x}^{2}dx=2{\pi }^2/3$. Comparison yieldsThe original sine series is not uniformly convergent on the whole open interval; that stronger assertion is unnecessary for the justified integration.

Let $B=|b|>0$. For $a>0$, integrate $e^{iaz}/(z^2+B^2)$ over the upper semicircle. Jordan's lemma removes the arc, and the residue theorem at $z=iB$ givesTaking the real part and using evenness in both $x$ and $a$ yieldsThe case $a=0$ follows directly from an inverse tangent primitive. The complex exponential is essential on the contour: the cosine alone has a growing exponential component.

For $a>0$, differentiation under the integral sign gives $-I_1^{\prime }(a)={\int }_0^{\infty }x\sin (ax)/(x^2+B^2)dx$. This derivative is a conditionally convergent improper integral; the Dirichlet test makes its tails uniform on compact subintervals of $a>0$, justifying the differentiation. Setting $a=1$ givesFinally let $J(a)={\int }_0^{\infty }\sin (ax)/[x(x^2+B^2)]dx$. Its differentiated integrand is absolutely integrable, so $J^{\prime }(a)=I_1(a)$ and $J(0)=0$. HenceIf $b=0$, $I_1$ and $I_3$ diverge at the origin, whereas $I_2={\int }_0^{\infty }\sin x/xdx=\pi /2$ is the corresponding Dirichlet integral.

For real $p>0$, substitute $t=px$ in the Laplace transform:The Gamma function integral is finite because $a>0$. For complex $p$, absolute convergence holds exactly when $\mathrm{Re}p>0$: the modulus of the integrand is $x^{a-1}{e}^{-(\mathrm{Re}p)x}$. On this half-plane the same answer holds with the branch $p^{-a}=\exp (-a\mathrm{Log}p)$ whose logarithm is real on the positive axis, by holomorphic continuation of the two sides.

If “exists” permits conditionally convergent ordinary improper integrals, there are additional boundary points when $0<a<1$: for $p=i\omega \ne 0$, the Dirichlet test gives convergence, with value $\Gamma (a)|\omega {|}^{-a}{e}^{-i\mathrm{sgn}(\omega )\pi a/2}$. For $a\ge 1$ these boundary integrals do not converge in that sense. At $p=0$, and for $\mathrm{Re}p<0$, the integral diverges. Thus $\mathrm{Re}p>0$ is the region of absolute convergence of a Laplace transform, with the stated conditional boundary qualification.