A power series \(\sum_na_nz^n\) has radius of convergence \(R\) given by \(1/R=\limsup_n\abs{a_n}^{1/n}\) (Cauchy–Hadamard); it converges absolutely for \(\abs{z}<R\), uniformly on compact subsets of \(\{\abs{z}<R\}\), and diverges for \(\abs{z}>R\). On the circle \(\abs{z}=R\) the behaviour depends on the coefficients. If \(\sum_n\abs{a_n}R^n<\infty\), the series converges absolutely at every point of the circle. By Dirichlet’s test, if \(R=1\) and \(a_n\) decreases to \(0\), the series converges at every point of \(\abs{z}=1\) except possibly \(z=1\). The series \(\sum z^n\), \(\sum z^n/n\), and \(\sum z^n/n^2\) converge at no point, at every point except \(z=1\), and at every point of \(\abs{z}=1\), respectively. Abel’s theorem: if \(\sum_na_n\) converges, then \(\lim_{r\to1^-}\sum_na_nr^n=\sum_na_n\).
Radius of convergence by the root test (Tag PR-NQ64T)
Let \((c_k)_{k\ge0}\) be complex numbers, and define \(R\in[0,\infty]\) by \[ \frac1R\coloneqq\limsup_{k\to\infty}\abs{c_k}^{1/k}, \] with the conventions \(1/0=\infty\) and \(1/\infty=0\). Then the power series \(f(z)=\sum_{k\ge0}c_kz^k\) converges absolutely for \(\abs{z}<R\), uniformly on every closed disc \(\abs{z}\le r\) with \(r<R\), and diverges for \(\abs{z}>R\). So \(R\) is the radius of convergence of the series.
Moreover \(f\) is holomorphic on \(D_R\coloneqq\ts{z\st\abs{z}<R}\), and it can be differentiated term by term: \[ f'(z)=\sum_{k\ge1}kc_kz^{k-1},\qquad \abs{z}<R, \] where the differentiated series has the same radius of convergence \(R\).
Root test for power series (Tag PR-TOK44)
Let \((a_n)_{n\ge0}\) be complex numbers, \(z_0\in\CC\), and \[ R\coloneqq\liminf_{n\to\infty}\abs{a_n}^{-1/n}=\frac{1}{\limsup_{n\to\infty}\abs{a_n}^{1/n}}\in[0,\infty], \] with the conventions \(1/0=\infty\) and \(1/\infty=0\). Then the power series \(\sum_{n\ge0}a_n(z-z_0)^n\)
converges absolutely on \(\ts{z\st\abs{z-z_0}<R}\),
converges uniformly on \(\ts{z\st\abs{z-z_0}\le r}\) for every \(r<R\), and
diverges on \(\ts{z\st\abs{z-z_0}>R}\).
Ratio test (Tag PR-WIMIM)
Let \((a_k)_{k\ge0}\) be nonzero complex numbers such that \(L\coloneqq\lim_{k\to\infty}\abs{\frac{a_{k+1}}{a_k}}\) exists in \([0,\infty]\).
If \(L<1\), then \(\sum_k a_k\) converges absolutely.
If \(L>1\), then \(\sum_k a_k\) diverges.
If \(L=1\), the test gives no information: \(\sum_k 1/k\) diverges and \(\sum_k 1/k^2\) converges, and both have \(L=1\).
In particular, if \(c_k\neq0\) and \(\lim_k\abs{c_{k+1}/c_k}=\ell\in[0,\infty]\), applying the test to \(a_k=c_kz^k\) shows that \(\sum_kc_kz^k\) has radius of convergence \(1/\ell\), with \(1/0=\infty\) and \(1/\infty=0\).
The $p$-test (Tag PR-5WUAP)
For \(p\in\RR\), the series \(\sum_{n\ge1} n^{-p}\) converges if and only if \(p>1\).
If \(p\le0\), then \(n^{-p}\ge1\) for all \(n\), so the terms do not tend to \(0\) and the series diverges. If \(p>0\), the function \(x\mapsto x^{-p}\) is positive and decreasing on \([1,\infty)\), so by the integral test the series converges if and only if \(\int_1^\infty x^{-p}\dx\) is finite, which holds if and only if \(p>1\).
Weierstrass M-test (Tag T-AZSCN)
Let \(\Omega\) be a set, let \(f_n\colon\Omega\to\CC\) for \(n\in\NN\), and suppose there are real numbers \(M_n\) with \[ \sup_{x\in\Omega}\abs{f_n(x)}\le M_n\quad\text{for all }n,\qquad\sum_{n\in\NN}M_n<\infty. \] Then \(f(x)\coloneqq\sum_{n\in\NN}f_n(x)\) converges absolutely and uniformly on \(\Omega\). If moreover \(\Omega\) is a topological space and every \(f_n\) is continuous, then \(f\) is continuous, by the uniform limit theorem.
For each \(x\), \(\sum_n\abs{f_n(x)}\le\sum_nM_n<\infty\), so the series converges absolutely. For \(N\ge1\) and all \(x\in\Omega\), \[ \abs{f(x)-\sum_{n\le N}f_n(x)}\le\sum_{n>N}M_n, \] and the right side is independent of \(x\) and tends to \(0\) as \(N\to\infty\).
Summation by parts (Tag PR-2MTPE)
Let \((a_k)_{k\ge1}\) and \((b_k)_{k\ge1}\) be sequences of complex numbers, and put \(A_n \coloneqq \sum_{k=1}^n a_k\) and \(B_n\coloneqq\sum_{k=1}^n b_k\) for \(n\ge1\), with \(A_0 \coloneqq 0\) and \(B_0\coloneqq0\).
For all integers \(1\le m\le n\), \[ \sum_{k=m}^n a_k b_k = A_nb_n - A_{m-1} b_m - \sum_{k=m}^{n-1} A_k(b_{k+1} - b_{k}). \]
For all sequences \((f_k)\) and \((g_k)\) of complex numbers and all integers \(m\le n\), writing \(\Delta g_k \coloneqq g_{k+1} - g_k\) and \(\Delta f_k\coloneqq f_{k+1}-f_k\), \[ \sum_{k=m}^n f_k \, \Delta g_k = f_{n+1} g_{n+1} - f_m g_m - \sum_{k=m}^n g_{k+1} \, \Delta f_k. \]
For every \(n\ge1\), \[ A_{n} B_{n}=\sum_{k=1}^{n} A_{k} b_{k}+\sum_{k=1}^{n} a_{k} B_{k-1}. \]
For (b), \(f_{k+1}g_{k+1}-f_kg_k=f_k\,\Delta g_k+g_{k+1}\,\Delta f_k\) for every \(k\); summing over \(m\le k\le n\) telescopes the left side to \(f_{n+1}g_{n+1}-f_mg_m\).
For (a), substitute \(a_k=A_k-A_{k-1}\): \[ \sum_{k=m}^n a_kb_k=\sum_{k=m}^nA_kb_k-\sum_{k=m-1}^{n-1}A_kb_{k+1}=A_nb_n-A_{m-1}b_m-\sum_{k=m}^{n-1}A_k(b_{k+1}-b_k). \]
For (c), \(A_kB_k-A_{k-1}B_{k-1}=A_k(B_{k-1}+b_k)-(A_k-a_k)B_{k-1}=A_kb_k+a_kB_{k-1}\) for every \(k\ge1\); summing over \(1\le k\le n\) and using \(A_0B_0=0\) gives the identity.
Identity (a) is the discrete analogue of integration by parts: for \(F\) an antiderivative of \(f\) on \([a,b]\) and \(g\) continuously differentiable, \[ \int_a^b f g = F(b)g(b) - F(a)g(a) - \int_a^b Fg'. \] The partial sums \(A_k\) play the role of \(F\), and the differences \(b_{k+1}-b_k\) the role of \(g'\).
Dirichlet's test (Tag FT-3ZL25)
Let \((a_n)_{n\ge1}\) be a sequence of real numbers and \((b_n)_{n\ge1}\) a sequence of complex numbers such that
\(a_{n+1} \leq a_n\) for all \(n\) and \(a_n \to 0\), and
there exists \(M\ge0\) such that \(\abs{\sum_{n=1}^N b_n} \leq M\) for all \(N\ge1\).
Then the series \(\sum_{n=1}^\infty a_n b_n\) converges.
Abel's test for power series on the unit circle (Tag L-MYZOX)
Let \((c_k)_{k\ge0}\) be a nonincreasing sequence of nonnegative real numbers with \(c_k\to0\), and let \(f(z)\coloneqq\sum_{k\ge0}c_kz^k\). Then the series \(f(z)\) converges for every \(z\) with \(\abs{z}=1\) and \(z\neq1\).
Let \(\abs{z}=1\) with \(z\neq1\). The partial sums of \(\sum_k z^k\) satisfy \(\abs{\sum_{k=0}^{N}z^k}=\abs{\frac{1-z^{N+1}}{1-z}}\le\frac{2}{\abs{1-z}}\) for every \(N\), so they are bounded. Since \((c_k)\) is nonincreasing with limit \(0\), Dirichlet's test applied to \(a_k=c_k\) and \(b_k=z^k\) shows that \(\sum_k c_kz^k\) converges.
Abel's theorem (Tag T-B7YTE)
Let \((c_k)_{k\ge0}\) be complex numbers such that \(\sum_{k\ge0}c_k\) converges. Then \(\sum_{k\ge0}c_kz^k\) converges for \(\abs z<1\), and \[ \lim_{r\to1^-}\sum_{k\ge0}c_kr^k=\sum_{k\ge0}c_k, \] where the limit is taken over real \(r\in(0,1)\).
Uniform convergence of a series through its tails (Tag PR-4LISY)
Let \(X\) be a set and let \(f_n\colon X\to\CC\) for \(n\ge1\) be functions such that \(\sum_{n=1}^\infty f_n(x)\) converges for every \(x\in X\). Then \(\sum_{n=1}^\infty f_n\) converges uniformly on \(X\) if and only if \[ \lim_{n\to \infty} \sup_{x\in X}\abs{ \sum_{k\geq n} f_k(x) } = 0. \]
Let \(S(x)\coloneqq\sum_{k\ge1}f_k(x)\) and \(S_n(x)\coloneqq\sum_{k=1}^nf_k(x)\). Then \(\sum_{k\ge n}f_k(x)=S(x)-S_{n-1}(x)\), so the displayed condition says exactly that \(\sup_{x\in X}\abs{S(x)-S_{n-1}(x)}\to0\), which is uniform convergence of the partial sums \(S_n\) to \(S\).
A power series of radius $1$ converging at every point of $S^1$ (Tag FE-VWNUI)
The power series \(\sum_{n=1}^\infty \frac{z^n}{n^2}\) has radius of convergence \(1\) and converges at every \(z\) with \(\abs{z}=1\). Its radius is \(1\) because \(\limsup_n(1/n^2)^{1/n}=1\). For \(\abs{z}=1\), \(\abs{z^n/n^2}=1/n^2\) and \(\sum_n 1/n^2<\infty\), so the series converges absolutely.
A power series of radius $1$ converging on $S^1$ except at $z=1$ (Tag FE-LAN3V)
The power series \(\sum_{n=1}^\infty \frac{z^n}{n}\) has radius of convergence \(1\), diverges at \(z=1\), and converges at every other point of \(S^1\). Its radius is \(1\) because \(\limsup_n(1/n)^{1/n}=1\). At \(z=1\) it is the harmonic series, which diverges. For \(\abs{z}=1\) with \(z\neq1\), the partial sums satisfy \(\abs{\sum_{n=1}^N z^n}=\abs{\frac{z(1-z^N)}{1-z}}\le\frac{2}{\abs{1-z}}\), and \(1/n\) decreases to \(0\), so the series converges by Dirichlet’s test.
A power series of radius $1$ diverging at every point of $S^1$ (Tag FE-EUOB2)
The power series \(\sum_{n=1}^\infty nz^n\) has radius of convergence \(1\) and diverges at every \(z\) with \(\abs{z}=1\). Its radius is \(1\) because \(\limsup_n n^{1/n}=1\). For \(\abs{z}=1\) the terms satisfy \(\abs{nz^n}=n\not\to0\), so the series diverges.