Series

A series is the formal or limiting sum of the terms of a sequence.

For a series ${\sum }_n{a}_n$, let $r={\mathrm{limsup}}_{n\to \infty }|a_n{|}^{1/n}$. If $r<1$, choose $r<\rho <1$; eventually $|a_n|\le {\rho }^n$, so geometric comparison gives absolute convergence. If $r>1$, the terms fail to tend to zero along a subsequence and the series diverges. At $r=1$, the test is inconclusive.

A telescoping series has partial sums in which consecutive terms cancel, commonly because its summand has the form $b_n-b_{n+1}$.

If the terms $a_n$ of a series do not converge to zero, then ${\sum }_n{a}_n$ diverges. This is the contrapositive of $a_n=S_n-S_{n-1}\to 0$ for convergent partial sums.

An infinite product is the limit of the finite partial products ${\prod }_{n=1}^N{a}_n$. If $a_n=1+u_n$, every factor is nonzero, and ${\sum }_n|u_n|<\infty$, then the product converges to a nonzero value.

The Weierstrass elementary factor has a simple zero at $w=1$ and no other zeros. Its logarithmic series begins at degree $p+1$: $\log E_p(w)=-{\sum }_{\nu \ge p+1}{w}^{\nu }/\nu$ for $|w|<1$. Raising the factor order $p$ makes the tail arbitrarily small on a fixed compact set and enables infinite product convergence from logarithmic tails for unrestricted discrete zero data.

If a tail of holomorphic factors admits logarithms whose absolute values have a summable bound uniformly on every compact set, their product converges locally uniformly to the exponential of the logarithmic sum. The tail is holomorphic and nowhere zero. Finite initial factors may have zeros; those finite factors determine the locations and orders of all zeros near any given compact set. The scalar sufficient condition ${\sum }_j|u_j|<\infty$ for factors $1+u_j$ follows from $|\log (1+u_j)|\le 2|u_j|$ for $|u_j|\le 1/2$.

For $a_j\ge 0$, the finite product satisfiesThe lower inequality follows by product expansion; the upper one uses $1+t\le e^t$. Since both the partial sums and partial products are increasing, one converges to a finite limit exactly when the other does. The monotone bounded sequence theorem proves both implications. The product cannot tend to zero in this nonnegative setting.

This infinite product converges uniformly on compact subsets of the complex plane. Pair the factors at $iz$ and $-iz$ in the Weierstrass product for the reciprocal gamma function, and use the Gamma reflection formula together with $\Gamma (1+w)=w\Gamma (w)$. This gives $[\Gamma (1+iz)\Gamma (1-iz){]}^{-1}=\sinh (\pi z)/(\pi z)$ and the displayed product. The removable value at $z=0$ is one. Matching zeros alone would not exclude multiplication by a nonvanishing entire function; the normalized product fixes that ambiguity.

A lattice sum is a series indexed by the points of a lattice. For example, comparison with annular lattice-point counts shows that ${\sum }_{(m,n)\ne (0,0)}(m^2+n^2{)}^{-s}$ converges when $s>1$.

For terms $(a_n)$, the $n$th partial sum is $S_n={\sum }_{j=1}^n{a}_j$. A series converges when its sequence of partial sums converges.

Let $(b_n)$ be positive and increase to infinity. If the series ${\sum }_{n\ge 1}{x}_n/b_n$ converges, thenThis follows from summation by parts: write $x_k/b_k$ as a convergent series and average its tails with the increasing weights $b_k$.

The harmonic series diverges. Grouping terms from $2^k+1$ through $2^{k+1}$ gives at least $2^k/2^{k+1}=1/2$ from every block.

If $(a_n)$ is a nonnegative decreasing sequence, then ${\sum }_{n\ge 1}{a}_n$ converges if and only if the condensed series ${\sum }_{k\ge 0}{2}^k{a}_{2^k}$ converges. Comparing each dyadic block with its first and last terms proves both directions.