In mathematics, the Weierstrass M-test is a test for determining whether an infinite series of functions converges uniformly and absolutely. It applies to series whose terms are bounded functions with real or complex values, and is analogous to the comparison test for determining the convergence of series of real or complex numbers. It is named after the German mathematician Karl Weierstrass (1815-1897).
Weierstrass M-test. Suppose that (fn) is a sequence of real- or complex-valued functions defined on a set A, and that there is a sequence of non-negative numbers (Mn) satisfying the conditions
|fn(x)|\leqMn
n\geq1
x\inA
infty | |
\sum | |
n=1 |
Mn
infty | |
\sum | |
n=1 |
fn(x)
A series satisfying the hypothesis is called normally convergent. The result is often used in combination with the uniform limit theorem. Together they say that if, in addition to the above conditions, the set A is a topological space and the functions fn are continuous on A, then the series converges to a continuous function.
Consider the sequence of functions
Sn(x)=
n | |
\sum | |
k=1 |
fk(x).
Since the series
infty | |
\sum | |
n=1 |
Mn
\forall\varepsilon>0:\existsN:\forallm>n>N:
m | |
\sum | |
k=n+1 |
Mk<\varepsilon.
\forallx\inA:\forallm>n>N
\left|Sm(x)-Sn
m | |
(x)\right|=\left|\sum | |
k=n+1 |
fk(x)\right|\overset{(1)}{\leq}
m | |
\sum | |
k=n+1 |
|fk(x)|\leq
m | |
\sum | |
k=n+1 |
Mk<\varepsilon.
The sequence is thus a Cauchy sequence in R or C, and by completeness, it converges to some number that depends on x. For n > N we can write
\left|S(x)-Sn(x)\right|=\left|\limm\toinftySm(x)-Sn(x)\right|=\limm\toinfty\left|Sm(x)-Sn(x)\right|\leq\varepsilon.
infty | |
\sum | |
k=1 |
fk(x)
Analogously, one can prove that
infty | |
\sum | |
k=1 |
|fk(x)|
A more general version of the Weierstrass M-test holds if the common codomain of the functions (fn) is a Banach space, in which case the premise
|fn(x)|\leqMn
is to be replaced by
\|fn(x)\|\leqMn
where
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