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The test works because the space of real numbers and the space of complex numbers (with the metric given by the absolute value) are both complete. From here, the series is convergent if and only if the partial sums:= = are a Cauchy sequence.
For example, the sum of 1/n where n has at most one 9, is a convergent series. But the sum of 1/n where n has no 9 is convergent. Therefore, the sum of 1/n where n has exactly one 9, is also convergent. Baillie [11] showed that the sum of this last series is about 23.04428 70807 47848 31968.
is used for the series, and, if it is convergent, to its sum. This convention is similar to that which is used for addition: a + b denotes the operation of adding a and b as well as the result of this addition, which is called the sum of a and b. Any series that is not convergent is said to be divergent or to diverge.
The series can be compared to an integral to establish convergence or divergence. Let : [,) + be a non-negative and monotonically decreasing function such that () =.If = <, then the series converges.
Otherwise, any series of real numbers or complex numbers that converges but does not converge absolutely is conditionally convergent. Any conditionally convergent sum of real numbers can be rearranged to yield any other real number as a limit, or to diverge. These claims are the content of the Riemann series theorem. [31] [32] [33]
More generally one can define summation methods slightly stronger than Borel's by taking the numbers b n to be slightly larger, for example b n = cnlog n or b n =cnlog n log log n. In practice this generalization is of little use, as there are almost no natural examples of series summable by this method that cannot also be summed by Borel's method.
Notably, these series provide examples of infinite sums that converge or diverge arbitrarily slowly. For instance, in the case of k = 2 {\displaystyle k=2} and α = 1 {\displaystyle \alpha =1} , the partial sum exceeds 10 only after 10 10 100 {\displaystyle 10^{10^{100}}} (a googolplex ) terms; yet the series diverges nevertheless.
For instance, for Alcuin's version of the problem, =: a camel can carry 30 measures of grain and can travel one leuca while eating a single measure, where a leuca is a unit of distance roughly equal to 2.3 kilometres (1.4 mi). The problem has =: there are 90 measures of grain, enough to supply three trips.