The Dirichlet Test
Theorem (The Dirichlet Test)
Let
X
be a metric space. If the functions
f
n
:
X
→
C
,
g
n
:
X
→
IR
,
n
∈
IN
obey
◦
F
n
(
x
) =
n
∑
m
=1
f
m
(
x
)
is bounded uniformly in
n
and
x
◦
g
n
+1
(
x
)
≤
g
n
(
x
)
for all
x
∈
X
and
n
∈
◦
b
g
n
(
x
)
B
n
∈
converges uniformly to zero on
X
then
∞
∑
n
=1
f
n
(
x
)
g
n
(
x
)
converges uniformly on
X
.
Proof:
The trick to this proof is the summation by parts formula, which we now derive.
s
n
(
x
) =
n
s
k
=1
f
k
(
x
)
g
k
(
x
)
=
F
1
(
x
)
g
1
(
x
) +
n
s
k
=2
[
F
k
(
x
)
−
F
k

1
(
x
)]
g
k
(
x
)
=
F
1
(
x
)
g
1
(
x
) +
n
s
k
=2
F
k
(
x
)
g
k
(
x
)
−
n
s
k
=2
F
k

1
(
x
)
g
k
(
x
)
=
n
s
k
=1
F
k
(
x
)
g
k
(
x
)
−
n

1
s
k
=1
F
k
(
x
)
g
k
+1
(
x
)
=
n
s
k
=1
F
k
(
x
)
±
g
k
(
x
)
−
g
k
+1
(
x
)
²
+
g
n
+1
(
x
)
F
n
(
x
)
So if
m > n
, the diFerence between the
m
th
and
n
th
partial sums is
s
m
(
x
)
−
s
n
(
x
) =
m
s
k
=
n
+1
F
k
(
x
)
±
g
k
(
x
)
−
g
k
+1
(
x
)
²
+
g
m
+1
(
x
)
F
m
(
x
)
−
g
n
+1
(
x
)
F
n
(
x
)
If
M
= sup
b

F
n
(
x
)

v
v
x
∈
X, n
∈
B
,
v
v
s
m
(
x
)
−
s
n
(
x
)
v
v
≤
M
m
s
k
=
n
+1
±
g
k
(
x
)
−
g
k
+1
(
x
)
²
+
Mg
m
+1
(
x
) +
n
+1
(
x
)
=
M
±
g
n
+1
(
x
)
−
g
m
+1
(
x
)
²
+
m
+1
(
x
) +
n
+1
(
x
)
= 2
n
+1
(
x
)
(1)
since
g
m
+1
(
x
)
≥
0,
g
n
+1
(
x
)
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 Winter '08
 JoelFeldman
 Math, Calculus, Mathematical Series, FN, dirichlet test

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