3.5 Series

Example

Given \(a_n=\frac{1}{2^n}\), then compute \(\sum_{k=0}^{\infty}\frac{1}{2^{k}}\)

\(s_0=a_0=1\), \(s_1=1+\frac{1}{2}=\sum^1_{k=0}\frac{1}{2^k}\), ... ,

\(s_{n}=1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n}}=\sum_{k=0}^{n}\frac{1}{2^{k}}\),

\(s_{n+1}=1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n}}+\frac{1}{2^{n+1}}=\sum _{k=0}^{n+1}\frac{1}{2^{k}}\)

Then \(\frac{1}{2}s_{n}=\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{n+1}} =\sum_{k=0}^{n+1}\frac{1}{2^{k}}\) and \(s_{n+1}=s_{n}+\frac{1}{2^{n+1}}\)​

Compute \(s_{n+1}-\frac{1}{2}s_{n}=1\implies s_{n}=\frac{1-\frac{1}{2^{n+1}}}{1-\frac{1}{2}}\)

Then \(\sum_{k=0}^{\infty}\frac{1}{2^{k}}=\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{2^{k}} =\lim_{n\to\infty}\frac{1-\frac{1}{2^{n+1}}}{1-\frac{1}{2}}\)

Revision algebra of convergent series

1. If \sum_{n=1}^{\infty}\alpha^{n} is convergent, then a_{n}\xrightarrow{n\to\infty}0

2. If \sum_{n=1}^{\infty}a^{n}=A and \sum_{n=1}^{\infty}b^{n}=B are convergent, then \forall\alpha,...

Firstly, we need to ensure the series is convergent, that is the limit of series exists

Absolutely Convergent

We say that a series \(\sum_{n=1}^{\infty}a_{n}\) is absolutely convergent if \(\sum_{n=1}^{\infty}|a_{n}|\) is convergent

The series of \(a_n=(-1)^n\frac{1}{n}\) is convergent but not absolutely convergent.

Theorem

If \(\sum_{n=1}^{\infty}\alpha^{n}\) is absolutely convergent, then \(\sum_{n=1}^{\infty}\alpha^{n}\) is convergent.

Proof: See this

Cauchy Criterium for Series

A series \(\sum_{n=1}^{\infty}a_{n}\) converges iff \(\forall \varepsilon >0,\exists N\in \N\) such that \(\forall n>m\geq N,|a_{m+1}+\cdots+a_{n}|<\varepsilon\)

Proof

\(S_{n}=\sum_{k=0}^{n}a_{k},S_{m}=\sum_{k=0}^{m}a_{k}\)

Since series is convergent, then it is a Cauchy sequence, then \(\forall m,n\geq N,|S_n-S_m|<\varepsilon\)

Then \(|\sum_{k=0}^{n}a_{k}-\sum_{k=0}^{m}a_{k}|=\left|a_{m+1}+\cdots+a_{n}\right|<\varepsilon\)

​#Method tool: Cauchy Criterions to deal with unknown limit#​

Comparison test for series

Recall this (Key: monotonic of series)

Series of functions

Take \(a_n=\frac{1}{2^n}\) as an example

\(a_n(\frac{1}{2})=a_n=\frac{1}{2^n},n\geq0\) \(a_n(x)=x^n,n\geq0\)

Then \(S_{n}(x)=\sum_{k=0}^{n}a_{n}\left(x\right)=\sum_{k=0}^{n}x^{k}=1+x+x^{2}+\cdots+ x^{n}\)

Then \(\sum_{k=0}^{\infty}x^{k}=\lim_{n\to\infty}\sum_{k=0}^{n}x^{k}=\lim\frac{1-x^{n+1}}{1-x}\) called rational function (use formula of Geometric series)

Then if \(x\in (-1,1)\), the limit exists, but if not, the series is divergent and limit doesn't exists.

Note: We cannot do this for all \(x\)

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Definition

We define a series of functions as we define a series related to a sequence using partial sums:

Let \((fₙ)\) be a sequence of functions defined on a subset \(A ⊆ ℝ\)

We define the function \(Sₙ(x)\) given by partial sums: \(S_N(x) = \sum_{n=1}^N f_n(x)\)​

Then, \(Sₙ(x)\) is a function on \(A\).

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We define \(\sum_{n=1}^\infty f_n(x) := \lim_{N \to \infty} S_N(x)\)​

We say that the series of functions: \(\sum_{n=1}^\infty f_n(x)\) converges to a function \(f\) if the sequence of partial sums converges pointwise:

\[ f(x) = \lim_{N \to \infty} S_N(x) = \lim_{N \to \infty} \sum_{n=1}^N f_n(x) \]

That is the limit exists for all \(x ∈ A\)

In such a case, we write: \(\sum_{n=1}^\infty f_n(x) = f(x)\)

Note

The limit may not exist for some values in A and may exist for others.

Hence, we focus on the subsets of A where the series converges.

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Definition

If a sequence of partial sums (\(S_N(x)\)) converges uniformly on a set \(A\), we say that the series converges uniformly on \(A\).

Theorem

Let (\(f_n\)) be a sequence of continuous functions and suppose that the series \(\sum_{n=1}^{\infty} f_n(x)\) converges uniformly to a function \(f\) on \(A\). Then, \(f\) is continuous on \(A\).

Proof

We will use Theorem (Continuous Limit Theorem)

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Consider the sequence of partial sums \(S_n(x)=f_1(x)+...+f_n(x)\)

This is the sum of continuous functions thus \(S_n(x)\) is continuous

Then \(f\) is continuous since \(S_n\) is continuous and converges uniform to \(f\) by that theorem

Theorem (Term by term differentiability)

Let (\(f_n\)) be a sequence of differentiable functions on a closed interval \([a, b]\), and assume that the series \(\sum_{n=1}^{\infty}f_{n}^{\prime}(x)\) converges uniformly to a function \(g\) on \([a, b]\).

If there exists a point \(x_0 \in [a, b]\) where the series \(\sum_{n=1}^{\infty}f_{n}(x_{0})\) converges, then \(\sum_{n=1}^{\infty} f_n\) converges uniformly to a differentiable function \(f\) satisfying that \(f^{\prime}(x)=g(x)\) on \([a, b]\), that is \(\left(\sum_{n=1}^{\infty}f_{n}(x)\right)^{\prime}=\sum_{n=1}^{\infty}f_{n}^{\prime} (x)\)​

Proof

We will use The second version of Differentiable Limit Theorem and let \(f_n=S_n\)

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Consider the sequence of partial sums \(S_n(x)=f_1(x)+f_2(x)+...+f_n(x)\)

Then \(S_n(x)\) is differentiable on \([a,b]\) \(\forall n\geq 1\), thus \(S_{n}^{\prime}=f_{1}^{\prime}\left(x\right)+\ldots+f_{n}^{\prime}\left(x\right)= \sum_{k=0}^{n}f_{k}^{\prime}\left(x\right)\)

By hypothesis \(S'_n(x)\xrightarrow{u}g\) and as \(\exists x_0\in[a,b]\) such that \(\sum_{n=0}^{\infty}f_{n}\left(x_{0}\right)\) is convergent.

By the differentiability theorem: \(\sum_{n=1}^{\infty}f_{n}\left(x\right)=f\) and \(f\) is differentiable and \(f^{\prime}=\sum_{n=1}^{\infty}f_{n}^{\prime}\left(x\right)\)

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Why we cannot directly use algebra of derivative term by term

Because here is infinite partial sum, it is a limit. And we know the derivative is limit existing(definition)

Then we are going to prove the limit of limit, it's strange, we need these theorems.

Theorem (Cauchy Criterium for uniform convergence of series)

A series \(\sum_{n=1}^{\infty} f_n(x)\) converges uniformly on a subset \(A \subset \mathbb{R}\) if and only if for every \(\varepsilon > 0\) there exists \(N \in \mathbb{N}\) such that \(|f_n(x) + f_{n+1}(x) + \dots + f_m(x)| < \varepsilon\) for all \(n > m \geq N\) and all \(x \in A\).

Proof

We use Cauchy Criterion for Uniform Convergence

Then let \(S_n(x)\) be \(f_n(x)\), then we know \(S_n(x)\) converges uniformly. (definition)

Then easily...

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Now we describe another criterium that helps us to decide when a series converges uniformly.

Theorem (Weierstrass M-test)

Let \((f_n)\) be a sequence of functions defined on a subset \(A \subset \mathbb{R}\).

Let \((M_n)\) be a sequence (of numbers) such that \(|f_n(x)| \leq M_n \quad \forall n \in \mathbb{N}, \forall x \in A\)

If the numerical series \(\sum_{n=1}^{\infty} M_n\) converges, then \(\sum_{n=1}^{\infty} f_n\) converges uniformly on \(A\).

Proof

NTP uniformly convergence, Cauchy criterium of sequence is a tool since we don't know the limit #Method tool: Cauchy Criterions to deal with unknown limit#​

NTP: \(\forall \varepsilon > 0, \exists N \in \mathbb{N} \text{ s.t. } |f_{m+1}(x) + \dots + f_n(x)| < \varepsilon \quad \forall n \geq m \geq N\)

But \(|f_{m+1}(x)+\dots+f_{n}(x)|\) \(\leq M_{m+1} + \dots + M_n = \sum_{k=m+1}^n M_k = \sum_{k=1}^n M_k - \sum_{k=1}^m M_k\)​\(< \varepsilon \text{ if } m \geq N\)

Because \(\sum_{n=0}^\infty M_n \text{ is convergent.}\)

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Example

\(\sum_{n=0}^{\infty} x^n = \frac{1}{1-x} = f(x), \quad x \in (-1, 1)\) and \(\left( \sum_{n=0}^{\infty} x^n \right)' = \left( \frac{1}{1-x} \right)' = \frac{1}{(1-x)^2}\)

Then \(\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}\)

Example: \(\left( \sum_{n=0}^{\infty} \frac{x^n}{n!} \right)' = \sum_{n=1}^{\infty} \frac{n x^{n-1}}{n!} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x\)

Example: Compute \(\sum_{n=1}^{\infty}\frac{n}{3^{n-1}}=\sum_{n=1}^{\infty}n\left(\frac{1}{3}\right )^{n-1}=\frac{1}{(1-\frac{1}{3})^{2}}=\frac{9}{4}\)

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Exercise: Analyze the convergence of the series of functions \(\sum_{n=0}^{\infty} \frac{\cos(2^n x)}{2^n}\)

We know \(f_{n}(x)=\frac{\cos(2^nx)}{2^n}<\frac{1}{2^n}=M_n\) and \(\sum_{n=1}^{\infty} M_n\) converges

Thus \(\sum_{n=0}^{\infty} \frac{\cos(2^n x)}{2^n}\) converges uniformly

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