In general, no, there is no smallest divergent series---given any divergent series, you can construct a new one that also diverges, but more slowly. If she is discovered, she will be killed. $\begingroup$ I don't think it's quite a duplicate though, as there is a difference between smallest divergent sum and largest convergent sum. In a world divided by factions based on virtues, Tris learns she's Divergent and won't fit in. Veronica Roth's entire Divergent series of books is available together in this set, with a bonus booklet about the series. {\displaystyle {1 \over 1}+{1 \over 2}+{1 \over 4}+{1 \over 8}+{1 \over 16}+{1 \over 32}+\cdots =2.} $\endgroup$ – Nethesis Jan 19 '15 at 9:38
Yesterday I was explaining the mechanics of numeric series to a friend of mine. Consult Is there a slowest rate of divergence of a series? 48 Metascore. The reciprocals of powers of 2 produce a convergent series (so the set of powers of 2 is "small"): 1 1 + 1 2 + 1 4 + 1 8 + 1 16 + 1 32 + ⋯ = 2. In a futuristic society where the five factions: Dauntless, Erudite, Amity, Candor and Abnegation, are the basis of the whole function, Tris Prior discovers that she's Divergent, therefore does not fit into one faction. It's like the difference between smallest positive real number and largest non-positive real number. There is no "smallest" divergent series of positive values that sums to infinity because you can always just cut the terms in half and repeat each one twice. When she discovers a plot to destroy Divergents, Tris and the mysterious Four must find out what … When I was talking about the divergence of harmonic series $\sum_{n=1}^\infty\frac{1}{n}$ and the convergence of series $\sum_{n=1}^\infty\frac{1}{n^\alpha}$, for all $\alpha>1$ (Dirichlet's Theorem), she asked me if the harmonic series is the smallest divergent series that exists.