1
Uniform Convergence of Series: The Weierstrass M-Test
Def. ∑∞
𝒊=𝟏 𝑴𝒊 = 𝑴 ; where 𝑀𝑖 , 𝑀 ∈ ℝ, means given:
𝑆1 = 𝑀1
𝑆2 = 𝑀1 + 𝑀2
𝑆3 = 𝑀1 + 𝑀2 + 𝑀3
⋮
𝑆𝑛 = 𝑀1 + 𝑀2 + 𝑀3 + ⋯ + 𝑀𝑛
then 𝑀 = lim 𝑆𝑛 .
𝑛→∞
Def. 𝑺(𝒙) = ∑∞ 𝑛
𝒊=𝟏 𝒇𝒊 (𝒙) if given 𝑆𝑛 (𝑥 ) = ∑𝑖=1 𝑓𝑖 (𝑥),
𝑆1 (𝑥) = 𝑓1 (𝑥)
𝑆2 (𝑥) = 𝑓1 (𝑥) + 𝑓2 (𝑥)
𝑆3 (𝑥) = 𝑓1 (𝑥) + 𝑓2 (𝑥 ) + 𝑓3 (𝑥)
⋮
𝑆𝑛 (𝑥) = 𝑓1 (𝑥) + 𝑓2 (𝑥 ) + 𝑓3 (𝑥) + ⋯ + 𝑓𝑛 (𝑥)
then lim 𝑆𝑛 (𝑥) = 𝑆(𝑥) where this limit means pointwise convergence.
𝑛→∞
𝑥𝑖−1
Ex. Let 𝑓𝑖 (𝑥 ) = ; then
(𝑖−1)!
𝑥2 𝑥3 𝑥𝑛−1
𝑆𝑛 (𝑥 ) = ∑𝑛𝑖=1 𝑓𝑖 (𝑥) = 1 + 𝑥 + + + ⋯ + (𝑛−1)!
2! 3!
𝑥2 𝑥3 𝑥𝑛−1
𝑆(𝑥 ) = lim 𝑆𝑛 (𝑥 ) = 1 + 𝑥 + + + ⋯ + (𝑛−1)! + ⋯ = 𝑒 𝑥 .
𝑛→∞ 2! 3!
, 2
Def. We say ∑∞ 𝒊=𝟏 𝒇𝒊 (𝒙) converges uniformly to 𝑺(𝒙) if the sequence of
functions 𝑆𝑛 (𝑥) converges uniformly to 𝑆(𝑥).
Theorem (Weierstrass 𝑀 Test): Let {𝑓𝑛 (𝑥 )} be a sequence of functions on 𝐼.
Suppose that each 𝑓𝑛 (𝑥) is bounded on 𝐼, i.e. there exists real numbers 𝑀𝑛 such
that |𝑓𝑛 (𝑥)| ≤ 𝑀𝑛 for all 𝑥 ∈ 𝐼. If ∑∞ ∞
𝑛=1 𝑀𝑛 converges then ∑𝑛=1 𝑓𝑛 (𝑥)
converges uniformly on 𝐼.
Proof: We know that 𝑆𝑛 (𝑥 ) = ∑𝑛 𝑖=1 𝑓𝑖 (𝑥) converges uniformly to 𝑆(𝑥) if and
only if for all 𝜖 > 0 there exists an 𝑁 ∈ ℤ+ , such that for all 𝑥 ∈ 𝐼, if 𝑛, 𝑚 ≥ 𝑁
then |𝑆𝑚 (𝑥 ) − 𝑆𝑛 (𝑥 )| < 𝜖 (by the theorem we proved in the last section).
Assuming 𝑚 > 𝑛:
𝑆𝑚 (𝑥 ) − 𝑆𝑛 (𝑥 ) = ∑𝑚 𝑛 𝑚
𝑖=1 𝑓𝑖 (𝑥 ) − ∑𝑖=1 𝑓𝑖 (𝑥 ) = ∑𝑖=𝑛+1 𝑓𝑖 (𝑥 ).
So we need to force |𝑆𝑚 (𝑥 ) − 𝑆𝑛 (𝑥 )| = |∑𝑚
𝑖=𝑛+1 𝑓𝑖 (𝑥 )| < 𝜖 .
So if we can find a 𝑁 ∈ ℤ+ , such that for all 𝑥 ∈ 𝐼, if 𝑛, 𝑚 ≥ 𝑁 then
|∑𝑚
𝑖=𝑛+1 𝑓𝑖 (𝑥 )| < 𝜖 we will have proved ∑𝑛=1 𝑓𝑛 (𝑥) converges uniformly on 𝐼.
∞
Since ∑∞ +
𝑛=1 𝑀𝑛 converges we know given any 𝜖 > 0 there exists an 𝑁′ ∈ ℤ such
that 𝑚, 𝑛 ≥ 𝑁′ implies that
| ∑𝑚
𝑖=𝑛+1 𝑀𝑖 | = 𝑀𝑛+1 + 𝑀𝑛+2 + ⋯ + 𝑀𝑚 < 𝜖 .
Let 𝑁 = 𝑁′. Then we have by the triangle inequality:
|𝑆𝑚 (𝑥 ) − 𝑆𝑛 (𝑥 )| = |∑𝑚
𝑖=𝑛+1 𝑓𝑖 (𝑥 )| < |𝑓𝑛+1 (𝑥 )| + |𝑓𝑛+2 (𝑥 )| + ⋯ |𝑓𝑚 (𝑥 )|
≤ 𝑀𝑛+1 + 𝑀𝑛+2 + ⋯ + 𝑀𝑚 < 𝜖.
So ∑∞
𝑛=1 𝑓𝑛 (𝑥) converges uniformly on 𝐼.
Uniform Convergence of Series: The Weierstrass M-Test
Def. ∑∞
𝒊=𝟏 𝑴𝒊 = 𝑴 ; where 𝑀𝑖 , 𝑀 ∈ ℝ, means given:
𝑆1 = 𝑀1
𝑆2 = 𝑀1 + 𝑀2
𝑆3 = 𝑀1 + 𝑀2 + 𝑀3
⋮
𝑆𝑛 = 𝑀1 + 𝑀2 + 𝑀3 + ⋯ + 𝑀𝑛
then 𝑀 = lim 𝑆𝑛 .
𝑛→∞
Def. 𝑺(𝒙) = ∑∞ 𝑛
𝒊=𝟏 𝒇𝒊 (𝒙) if given 𝑆𝑛 (𝑥 ) = ∑𝑖=1 𝑓𝑖 (𝑥),
𝑆1 (𝑥) = 𝑓1 (𝑥)
𝑆2 (𝑥) = 𝑓1 (𝑥) + 𝑓2 (𝑥)
𝑆3 (𝑥) = 𝑓1 (𝑥) + 𝑓2 (𝑥 ) + 𝑓3 (𝑥)
⋮
𝑆𝑛 (𝑥) = 𝑓1 (𝑥) + 𝑓2 (𝑥 ) + 𝑓3 (𝑥) + ⋯ + 𝑓𝑛 (𝑥)
then lim 𝑆𝑛 (𝑥) = 𝑆(𝑥) where this limit means pointwise convergence.
𝑛→∞
𝑥𝑖−1
Ex. Let 𝑓𝑖 (𝑥 ) = ; then
(𝑖−1)!
𝑥2 𝑥3 𝑥𝑛−1
𝑆𝑛 (𝑥 ) = ∑𝑛𝑖=1 𝑓𝑖 (𝑥) = 1 + 𝑥 + + + ⋯ + (𝑛−1)!
2! 3!
𝑥2 𝑥3 𝑥𝑛−1
𝑆(𝑥 ) = lim 𝑆𝑛 (𝑥 ) = 1 + 𝑥 + + + ⋯ + (𝑛−1)! + ⋯ = 𝑒 𝑥 .
𝑛→∞ 2! 3!
, 2
Def. We say ∑∞ 𝒊=𝟏 𝒇𝒊 (𝒙) converges uniformly to 𝑺(𝒙) if the sequence of
functions 𝑆𝑛 (𝑥) converges uniformly to 𝑆(𝑥).
Theorem (Weierstrass 𝑀 Test): Let {𝑓𝑛 (𝑥 )} be a sequence of functions on 𝐼.
Suppose that each 𝑓𝑛 (𝑥) is bounded on 𝐼, i.e. there exists real numbers 𝑀𝑛 such
that |𝑓𝑛 (𝑥)| ≤ 𝑀𝑛 for all 𝑥 ∈ 𝐼. If ∑∞ ∞
𝑛=1 𝑀𝑛 converges then ∑𝑛=1 𝑓𝑛 (𝑥)
converges uniformly on 𝐼.
Proof: We know that 𝑆𝑛 (𝑥 ) = ∑𝑛 𝑖=1 𝑓𝑖 (𝑥) converges uniformly to 𝑆(𝑥) if and
only if for all 𝜖 > 0 there exists an 𝑁 ∈ ℤ+ , such that for all 𝑥 ∈ 𝐼, if 𝑛, 𝑚 ≥ 𝑁
then |𝑆𝑚 (𝑥 ) − 𝑆𝑛 (𝑥 )| < 𝜖 (by the theorem we proved in the last section).
Assuming 𝑚 > 𝑛:
𝑆𝑚 (𝑥 ) − 𝑆𝑛 (𝑥 ) = ∑𝑚 𝑛 𝑚
𝑖=1 𝑓𝑖 (𝑥 ) − ∑𝑖=1 𝑓𝑖 (𝑥 ) = ∑𝑖=𝑛+1 𝑓𝑖 (𝑥 ).
So we need to force |𝑆𝑚 (𝑥 ) − 𝑆𝑛 (𝑥 )| = |∑𝑚
𝑖=𝑛+1 𝑓𝑖 (𝑥 )| < 𝜖 .
So if we can find a 𝑁 ∈ ℤ+ , such that for all 𝑥 ∈ 𝐼, if 𝑛, 𝑚 ≥ 𝑁 then
|∑𝑚
𝑖=𝑛+1 𝑓𝑖 (𝑥 )| < 𝜖 we will have proved ∑𝑛=1 𝑓𝑛 (𝑥) converges uniformly on 𝐼.
∞
Since ∑∞ +
𝑛=1 𝑀𝑛 converges we know given any 𝜖 > 0 there exists an 𝑁′ ∈ ℤ such
that 𝑚, 𝑛 ≥ 𝑁′ implies that
| ∑𝑚
𝑖=𝑛+1 𝑀𝑖 | = 𝑀𝑛+1 + 𝑀𝑛+2 + ⋯ + 𝑀𝑚 < 𝜖 .
Let 𝑁 = 𝑁′. Then we have by the triangle inequality:
|𝑆𝑚 (𝑥 ) − 𝑆𝑛 (𝑥 )| = |∑𝑚
𝑖=𝑛+1 𝑓𝑖 (𝑥 )| < |𝑓𝑛+1 (𝑥 )| + |𝑓𝑛+2 (𝑥 )| + ⋯ |𝑓𝑚 (𝑥 )|
≤ 𝑀𝑛+1 + 𝑀𝑛+2 + ⋯ + 𝑀𝑚 < 𝜖.
So ∑∞
𝑛=1 𝑓𝑛 (𝑥) converges uniformly on 𝐼.
