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If $f$ is a multiplicative function with $|f(n)|\le 1$, then for $\Re s>1$ its Dirichlet series has the Euler productIndeed, expanding the product over a finite set of primes and using unique prime factorization gives the sum over integers having no other prime factors. Moreover,so absolute convergence permits rearrangement and passage to the limit over all primes. This proves the formula.

Let $T_N={\sum }_{n=1}^N|a_n|$ and $S_N={\sum }_{n=1}^N{a}_n$. If $(T_N)$ converges, it is a Cauchy sequence. Thus, for every $\epsilon >0$, some $N$ makeswhere the triangle inequality gives the bound. Therefore $(S_N)$ is a Cauchy sequence in the complex numbers. These form a complete metric space, so $(S_N)$ has a finite limit of a sequence. Equivalently, apply the Cauchy criterion for series. Hence absolute convergence implies convergence.

For a series of complex numbers, form the partial sums $S_N={\sum }_{n=1}^N{a}_n$. The series converges when these partial sums have a finite limit of a sequence $S\in \mathbb{C}$:The series has absolute convergence when ${\sum }_{n=1}^{\infty }|a_n|$ converges as a series of nonnegative real numbers. A series beginning at $n=0$ uses the same definitions with that starting index.

If $z_0=0$, there is no $z$ with $|z|<|z_0|$, so the claim is vacuous. Suppose $z_0\ne 0$. Convergence of $\sum a_n{z}_0^n$ implies $a_n{z}_0^n\to 0$, because terms of a convergent series tend to zero. In particular this is a bounded sequence, so $|a_n{z}_0^n|\le C$ for some finite $C$ and every $n$.

For $|z|<|z_0|$, put $q=|z|/|z_0|<1$. ThenThe geometric series and comparison test for series proveThus convergence at one point gives absolute convergence at every point strictly closer to the centre.

The radius of convergence of a power series ${\sum }_{n=0}^{\infty }{a}_n{z}^n$ is the number $R\in [0,\infty ]$ such that the series has absolute convergence for $|z|<R$ and diverges for $|z|>R$. Equivalently,Use $r^0=1$, including at $r=0$. The supremum may be $+\infty$. There is no general conclusion at $|z|=R$: boundary convergence depends on the coefficients. If $R=0$, only the centre is guaranteed; if $R=\infty$, every finite $z$ lies inside the disk of convergence.

All three examples below have positive integer coefficients, so they work under either convention for whether $0\in \mathbb{N}$.

For $R=1$, take $\boxed{a_n=1}$. This is the geometric series: it converges for $|z|<1$, and its terms fail to tend to zero for $|z|\ge 1$.

For $R=0$, take $\boxed{a_n=n!}$, using the factorial and $0!=1$. For every fixed $z\ne 0$, the absolute values of successive terms have ratio $(n+1)|z|$. This eventually exceeds two, so the terms grow without bound and the series cannot converge. At $z=0$ it is finite, giving radius of convergence zero.

For $R=1/\sqrt{2}$, takeusing the floor function. The coefficients satisfyThe upper bound and the comparison test for series give absolute convergence when $\sqrt{2}|z|<1$. The lower bound prevents the terms tending to zero when $\sqrt{2}|z|\ge 1$. Hence the radius of convergence is exactly $1/\sqrt{2}$.

Take $0\le r<R$. At the point $r$, the power series has absolute convergence, so ${\sum }_{n=0}^{\infty }|a_n|r^n<\infty$. On the whole interval $[-r,r]$,The Weierstrass M-test with $M_n=|a_n|r^n$ proves uniform convergence on $[-r,r]$. It also gives the explicit tail estimateFor $r=0$ the interval consists only of zero, and all positive-degree terms vanish. If $R=\infty$, the same argument applies to every finite $r$; if $R=0$, there is no $r\ge 0$ with $r<R$.

The Weierstrass M-test states: if functions $f_n:E\to \mathbb{R}$ or $\mathbb{C}$ satisfythen $\sum f_n$ has uniform convergence and absolute convergence on $E$. In fact $\sum |f_n|$ also has uniform convergence, since every tail is bounded by the corresponding tail of $\sum M_n$:

For a nonnegative integer-valued random variable, its probability generating function isThe convention $s^0=1$, including at $s=0$, gives $G(0)=p(0)$. The power series has absolute convergence for every real $|s|\le 1$, since ${\sum }_{j}p(j)|s{|}^j\le 1$. More generally its real domain as a finite expected value is $\{s\in \mathbb{R}:{\sum }_{j}p(j)|s{|}^j<\infty \}$, which may extend beyond $[-1,1]$. No extension beyond that guaranteed interval is automatic.

First, the Fourier coefficients of the Fourier cosine series decay as $O(n^{-2})$, whereas those of the Fourier sine series are of order $n^{-1}$. In particular, the absolute convergence bound for the Fourier cosine series gives a uniform tail estimateThus its tail is absolutely controlled using fewer Fourier harmonics. The Fourier sine series does not have an absolutely summable sequence of Fourier coefficients.

Second, the even extension is continuous, so its Fourier cosine series converges uniformly on the closed interval and approximates the nonzero endpoint limits correctly. The odd periodic extension has jumps, producing the Gibbs phenomenon in its Fourier sine series near zero and $L$. Every Fourier sine series partial sum is zero at both endpoints, so by approaching $L$ from inside $(0,L)$ one obtainsHence the Fourier sine series approximation cannot converge uniformly even on the whole open interval, despite pointwise convergence at each interior point. The Fourier cosine series approximation has faster coefficient decay and avoids the persistent errors near the endpoints.

A real or complex sequence $(a_n)$ is a Cauchy sequence ifFor $z_n=x_n+i{y}_n$, the complex modulus givesThus the same $N$ works for both coordinate Cauchy sequences. Conversely, if both real coordinate Cauchy sequences converge to $x$ and $y$, then $|z_n-(x+iy)|\le |x_n-x|+|y_n-y|\to 0$. This proves the needed completeness for complex numbers from the assumed completeness of the real numbers.

Now suppose ${\sum }_{n\ge 0}|z_n|$ converges. Its real partial sums form a Cauchy sequence. Given $\epsilon >0$, sufficiently late tails therefore satisfy ${\sum }_{k=n+1}^m|z_k|<\epsilon$. For $s_m={\sum }_{k=0}^m{z}_k$, the triangle inequality yieldsThe complex partial sums are a Cauchy sequence, hence converge by the coordinate argument. Therefore absolute convergence implies convergence for a complex series.

Define the radius of convergence of the power series bywith $R=\infty$ when this set is unbounded. It is nonempty because the power series converges at zero.

If $|z|<R$, the definition of supremum supplies a point $w$ of convergence with $|z|<|w|$. The terms $a_n{w}^n$ of a convergent series tend to zero and are bounded, say $|a_n{w}^n|\le M$. ThereforeThe right-hand side has a convergent geometric series, since $|z|/|w|<1$. The comparison test proves absolute convergence at $z$.

If $|z|>R$ and the power series converged at $z$, then $|z|$ would belong to the set whose supremum is $R$, an immediate contradiction. ThusNo assertion about the boundary $|z|=R$ follows from this argument. If $R=0$ the first assertion is vacuous; if $R=\infty$ the second is vacuous.

Split the power series into even and odd indices:whenever the two series converge absolutely. If $|z|<1$, the even part is a convergent geometric series, while the odd part is bounded termwise in absolute value by ${\sum }_j|z{|}^{2j+1}$, also a convergent geometric series. Hence there is absolute convergence throughout $|z|<1$.

If $|z|>1$, the odd terms have absolute values $|z{|}^{2j+1}/(2j+1{)}^2$. Their successive ratio tends to $|z{|}^2>1$, so these terms grow eventually and do not tend to zero. The term test for divergence therefore forces divergence of the original power series. Consequently its radius of convergence is

A series ${\sum }_{n=1}^{\infty }{a}_n$ converges if its partial sums $s_N={\sum }_{n=1}^N{a}_n$ have a finite limit of a sequence $S$ as $N\to \infty$; its sum is then $S$. It has absolute convergence if the series ${\sum }_{n=1}^{\infty }|a_n|$ converges, where $|a_n|$ is the absolute value of the term.

The truncated Fourier cosine series gives the better approximation on the closed interval, in the senses of uniform convergence and the asymptotic L2 norm error. To make the comparison precise, let $C_N$ and $S_N$ retain the frequencies $1\le n\le N$, with the constant term also included in $C_N$.

The cosine Fourier coefficients are $O(n^{-2})$. In fact, absolute convergence gives the uniform tail boundFor the Fourier sine series, every Fourier partial sum satisfies $S_N(0)=S_N(\pi )=0$. Thusso there is no uniform convergence to $f$ on the closed interval. Its odd extension has jump discontinuities, producing the Gibbs phenomenon near the endpoints; its Fourier coefficients decrease only like $1/n$.

The advantage persists in the L2 norm, which ignores individual endpoint values. By Parseval's identity and Fourier orthogonality,The nonzero odd cosine coefficients and both parity classes of sine coefficients also give matching lower bounds of these orders by comparison with the corresponding p-series. Hence their L2 norm errors decrease respectively like $N^{-3/2}$ and $N^{-1/2}$. This is a comparison of approximation quality in specified norms, not a claim that one finite sum has smaller error at every individual interior point.

A series ${\sum }_{n=1}^{\infty }{a}_n$ converges when its partial sums $s_N={\sum }_{n=1}^N{a}_n$ have a finite limit of a sequence.

The comparison test says that if $0\le a_n\le b_n$ eventually and $\sum b_n$ converges, then $\sum a_n$ converges. Indeed, the partial sums of the nonnegative terms $a_n$ are increasing, and are bounded above by a fixed initial contribution plus the sum of the comparison series. The bounded monotone sequence theorem proves their convergence. Finite initial terms do not affect convergence. Applying this result to $|a_n|$ also proves absolute convergence whenever $|a_n|\le b_n$ eventually with $\sum b_n<\infty$.

The implication is true, with absolute convergence. Since $n^{3/2}{a}_n\to 0$, there is $N$ such that $|a_n|\le n^{-3/2}$ for $n\ge N$. The p-series $\sum n^{-3/2}$ converges because its exponent exceeds one. The comparison test proves $\sum |a_n|<\infty$, and absolute convergence implies convergence of $\sum a_n$.

Let $u(x)=(\log x)x^{-2/3}$. Its derivative iswhich is negative for $x>e^{3/2}$. Also $u(x)\to 0$, since a positive power eventually dominates the natural logarithm. Thus the alternating series test proves convergence of $\sum (-1{)}^{n}u(n)$.

For $n\ge 3$, $u(n)\ge n^{-2/3}$, whose p-series diverges. The comparison test therefore rules out absolute convergence. The answer is conditional convergence.

For $t\ge 0$, $|\sin t|\le t$, soThe integral test applies to the decreasing positive comparison function, andThe comparison test proves that the series converges; since all terms are nonnegative, this is also absolute convergence.

For a power series $\sum a_n{z}^n$, its radius of convergence $R\in [0,\infty ]$ separates absolute convergence for $|z|<R$ from divergence for $|z|>R$. One may define $R=\sup \{r\ge 0:\sum |a_n|r^n<\infty \}$, allowing $R=\infty$.

To justify convergence inside the radius, choose $|z|<r<R$ with $\sum |a_n|r^n<\infty$. In particular $|a_n|r^n\le M$ for some $M$, soThe comparison test with a geometric series proves absolute convergence. Conversely, convergence at a point of modulus $\rho$ makes $a_n{z}^n$ bounded there and gives absolute convergence at all smaller moduli by the same estimate. Consequently a convergent point of modulus greater than $R$ would contradict the definition of $R$.

For $0<R<\infty$, the real domain of convergence is one of $(-R,R)$, $[-R,R)$, $(-R,R]$, or $[-R,R]$. At $R=1$ these are realized respectively byThe endpoint assertions follow from the harmonic series, alternating series test, and p-series. Replacing $z$ by $z/R$ realizes every finite positive radius. The degenerate domain $\{0\}$ occurs for $\sum n!z^n$, while the entire real line occurs for $\sum z^n/n!$. Their radii are zero and infinity by the ratio test.