Infinite Sequences and Series

A finite sum of real numbers is always well-defined, but making sense of an infinite series needs the idea of convergence. Consider $\dfrac12+\dfrac14+\dfrac18+\cdots$ — can a single numerical value be assigned to this infinite sum?

The cake experiment. Put a whole cake on plate $A$, plate $B$ empty. Repeatedly: cut the cake remaining on $A$ into two equal halves, move one half to $B$, leave the other on $A$. At stage $n$: plate $A$ holds $\tfrac1{2^n}$, plate $B$ holds $\tfrac12+\tfrac14+\cdots+\tfrac1{2^n}$. Intuitively, "finally," nothing is left on $A$ (its amount "goes" to $0$) and the whole cake ends up on $B$ (whose running total "goes" to $1$) — i.e. $\tfrac12+\tfrac14+\tfrac18+\cdots$ "is" $1$.

Convergence, precisely. A sequence $(a_n)$ converges to a number $a$ — written $a_n\to a$, or $\lim_{n\to\infty}a_n=a$ — if, for any given (arbitrarily small) positive tolerance, there is a stage beyond which $a_n$ stays within that tolerance of $a$. Not every sequence converges: $1,0,1,0,1,0,\ldots$ never settles near a single value. But a sequence that does converge, converges to a unique limit — it cannot tend to two different numbers at once.

Definition 5.6 (series convergence). For a series $\sum_{n=1}^\infty a_n$, let $s_n=a_1+a_2+\cdots+a_n$ be its partial-sum sequence. If $(s_n)$ converges, say $s_n\to s$, the series is called convergent with sum $s$, written $\sum_{n=1}^\infty a_n=s$.

For the cake series, $s_n=\tfrac12+\tfrac14+\cdots+\tfrac1{2^n}=\tfrac{2^n-1}{2^n}\to1$, confirming $\tfrac12+\tfrac14+\tfrac18+\cdots=1$. …