11.11 Series

Some more criteria

Proposition

If \(a_{n},b_{n}>0\) and \(\lim_{n \to \infty} \frac{a_n}{b_n} = l \neq 0\), then \(\sum_{n=1}^{\infty} a_n < \infty\) if and only if \(\sum_{n=1}^{\infty} b_n < \infty\).

Proof

Hypothesis \(\frac{a_n}{b_n} \to l \neq 0\).

Since \(\frac{a_n}{b_n} > 0 \Rightarrow l > 0\), take \(\varepsilon = \frac{l}{2} \Rightarrow \exists n_0 : n > n_0 \Rightarrow \left| \frac{a_n}{b_n} - l \right| < \frac{l}{2}\)

Then \(\frac{l}{2}<\frac{a_{n}}{b_{n}}<\frac{3l}{2}\)

Since \(b_{n}\to0\Rightarrow\begin{cases}0<a_{n}<\frac{3l}{2}b_{n}\\ 0<b_{n}<\frac{2}{l}a_{n}\end{cases}\forall n>n_0\)

And Sand Lemma:

If \(\sum_{n=1}^{\infty} b_n < \infty \Rightarrow \sum_{n=1}^{\infty} a_n < \infty\).

If \(\sum_{n=1}^{\infty} a_n < \infty \Rightarrow \sum_{n=1}^{\infty} b_n < \infty\).

Root test (Cauchy)

If \(\sqrt[n]{|a_n|} \to l\) as \(n \to \infty\), then:

If \(l < 1\) then \(\sum a_n\) is absolutely convergent
If \(l > 1\) then \(\sum a_n\) is Divergent.
If \(l = 1\), inconclusive.

Proof: Since \(\sqrt[n]{|a_n|} \geq 0 \Rightarrow l \geq 0\).

If \(l < 1\)

Let \(\varepsilon > 0\) such that \(l + \varepsilon < 1\), \(\left(\varepsilon = \frac{1 - l}{2}\right)\)

There is an \(n_0\) such that for all \(n > n_0\), \(\left| \sqrt[n]{|a_n|} - l \right| < \varepsilon\).

\(\Rightarrow \sqrt[n]{|a_n|} < l + \varepsilon=\alpha<1\) \(\Rightarrow 0 \leq |a_n| < \alpha^n \quad \forall n > n_0\)

Sandwich: \(\Rightarrow \sum_{n=1}^{\infty} |a_n| < \infty \quad \text{since } \sum_{n=1}^{\infty} \alpha^n < \infty\)

If \(l > 1\), let \(\varepsilon\) such that \(l - \varepsilon > 1\) (\(\alpha=l-\varepsilon\)).

Since \(\sqrt[n]{|a_n|} \to l\), then \(\exists n_0 : n > n_0 \Rightarrow \left| \sqrt[n]{|a_n|} - l \right| < \varepsilon\)

\(\Rightarrow l - \varepsilon < \sqrt[n]{|a_n|}\) \(\Rightarrow1<\alpha<\sqrt[n]{|a_{n}|}\Rightarrow1<\alpha^{n}<|a_{n}|\)

\(\Rightarrow |a_n|\) is NOT Bounded Above since \(\alpha^n\) is not.

\(\Rightarrow a_n \nrightarrow 0 \Rightarrow \sum_{n=1}^{\infty} a_n\) is Divergent.

In tutorial we saw

\(\sqrt[n]{\frac{1}{n}} \to 1\) (Same as \(\sqrt[n]{n} \to 1\)), also \(\sqrt[n]{\frac{1}{n^2}} \to 1\)

But \(\sum \frac{1}{n} = \infty\), \(\sum \frac{1}{n^2} < \infty\)

Example

\(a_n = \alpha^n\), \(\alpha\) any.

Apply root test: \(\sqrt[n]{|a_n|} = \sqrt[n]{|\alpha^n|} = |\alpha|\)

If \(|\alpha| < 1\), \(\sum a_n\) is Absolutely convergent

If \(|\alpha| > 1\), \(\sum a_n\) is Divergent.

If \(|\alpha| = 1\), I don't know.

If \(\alpha = 1 \Rightarrow \sum 1^n = \infty\)

If \(\alpha = -1 \Rightarrow \sum (-1)^n\) goes \(-1/0\)

D'Alembert Criterion (test)

If \(\lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} = l\), then

If \(l < 1 \Rightarrow \sum a_n\) is Absolutely convergent
If \(l > 1 \Rightarrow \sum a_n\) is Divergent.
If \(l = 1 \Rightarrow\) inconclusive.

Proof

Suppose \(l < 1\).

Let \(\varepsilon > 0\) such that \(l + \varepsilon < 1\) (\(\alpha=l+\varepsilon\)).

\(\exists n_0:n>n_0\Rightarrow\left|\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}-l\right|<\varepsilon\) \(\Rightarrow \frac{|a_{n+1}|}{|a_n|} < l + \varepsilon = \alpha\)

Let \(N=n_0+1\Rightarrow|a_{N+1}|<\alpha|a_{N}|\)

\(n=N+1\Rightarrow|a_{N+2}|<\alpha|a_{N+1}|\)

\(n=N+2\Rightarrow|a_{N+3}|<\alpha|a_{N+2}|\)

\(\Rightarrow|a_{N+k}|<\alpha^{k}|a_{N}|,\forall k\geq1\)

\(\Rightarrow\sum_{k=1}^{\infty}|a_{N+k}|<\infty\quad(\text{since }\sum_{k=1}^{\infty}\alpha^{k}\left|a_{N}\right|<\infty)\quad(|a_N|\text{constant number})\)

\(\Rightarrow\sum_{n=N+1}^{\infty}|a_{n}|<\infty\) \(\Rightarrow\sum_{n=1}^{\infty}|a_{n}|<\infty\Rightarrow\sum_{n=1}^{\infty}a_{n}\text{ is absolutely convergent}\) * If \(l > 1\)

Let \(\varepsilon > 0\) such that \(l - \varepsilon = \alpha > 1\)

\(\exists n_0:\forall n>n_0\Rightarrow\left|\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}-l\right|<\varepsilon\) \(\Rightarrow\frac{|a_{n+1}|}{|a_{n}|}>l-\varepsilon=\alpha>1\Rightarrow\left|a_{n+1}\right|>\alpha\left|a_{n}\right|\)

Let \(N=n_0+1\Rightarrow|a_{N+1}|>\alpha|a_{N}|\)

\(n=N+1\Rightarrow|a_{N+2}|>\alpha|a_{N+1}|\)

\(n=N+2\Rightarrow|a_{N+3}|>\alpha|a_{N+2}|\)

\(\Rightarrow|a_{N+k}|>\alpha^{k}|a_{N}|,\forall k\geq1\)

\(\Rightarrow\sum_{k=1}^{\infty}|a_{N+k}|=\infty\quad(\text{since }\sum_{k=1}^{\infty}\alpha^{k}\left|a_{N}\right|=\infty)\quad(|a_N|\text{constant number})\)

\(\Rightarrow\sum_{n=N+1}^{\infty}|a_{n}|=\infty\) \(\Rightarrow\sum_{n=1}^{\infty}|a_{n}|=\infty\Rightarrow\sum_{n=1}^{\infty}a_{n}\text{ is divergent}\) * \(a_n = \frac{1}{n}\), \(\frac{a_{n+1}}{a_n} = \frac{n}{n+1} \to 1\) (Divergent)

\(a_n = \frac{1}{n^2}\), \(\frac{a_{n+1}}{a_n} = \frac{n^2}{(n+1)^2} = \frac{n^2}{n^2 + 2n + 1} \to 1\) (A.C.)

Example

We know that if \(p\) and \(q\) are polynomials, and \(a_n = p(n)\), \(b_n = q(n)\).

What is \(\lim_{n \to \infty} \frac{a_n}{b_n} =\)\(\begin{cases}0 & \text{if }\deg(q)>\deg(p)\\ \frac{\text{lead}(p)}{\text{lead}(q)} & \text{if }\deg(q)=\deg(p)\\ \pm\infty & \text{if }\deg(q)<\deg(p)\end{cases}\)

\(\sum_{n=1}^{\infty} p(n) \alpha^n\) \(a_n = \frac{n^4 + 3n - 4}{2^n}\) ? \(p(n) = n^4 + 3n - 4\), \(\alpha = \frac{1}{2}\)

\(\frac{a_{n+1}}{a_n} = \frac{p(n+1) \ \alpha^{n+1}}{p(n) \ \alpha^n} = \frac{p(n+1)}{p(n)} \alpha \to \alpha\)

Therefore: if \(|\alpha| < 1 \Rightarrow \sum p(n) \alpha^n\) is A.C.

if \(|\alpha| > 1 \Rightarrow \sum p(n) \alpha^n\) is Divergent.
Let \(x \in \mathbb{R}\), \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\) is A.C. \(\forall x\).

D'Alembert: \(\frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}} = \frac{n! \ x^{n+1}}{(n+1)! \ x^n} = \frac{1}{n+1} \cdot x\) \(\to l = 0\)

Product of series

What is the "distributivity" of \(\left( \sum_{n=0}^{\infty} a_n \right) \left( \sum_{n=0}^{\infty} b_n \right)\)?

Consider
\((a_0+a_1+a_2+a_3+\dots)(b_0+b_1+b_2+\dots)=(a_0 b_0) + (a_0 b_1 + a_1 b_0) + (a_0 b_2 + a_1 b_1 + a_2 b_0) + \dots\)

Let us define \(c_n = \sum_{i=0}^{n} a_i b_{n-i}\), then \(c_0 = a_0 b_0\), \(c_1 = (a_0 b_1 + a_1 b_0)\), \(c_2 = (a_0 b_2 + a_1 b_1 + a_2 b_0)\)

Thus \(\sum_{n=0}^{\infty}c_{n}=\left(\sum_{n=0}^{\infty}a_{n}\right)\left(\sum_{n=0}^{\infty}b_{n}\right)\) where \(c_n = \sum_{i=0}^{n} a_i b_{n-i}\),

Theorem

Note: \(c_n = \sum_{k=0}^n a_k b_{n-k}\) (The Cauchy product) and \(\sum_{n=0}^{\infty}c_{n}=\left(\sum_{n=0}^{\infty}a_{n}\right)\left(\sum_{n=0}^{\infty}b_{n}\right)\)

At least one of these series converges absolutely

If \(\sum_{n=0}^{\infty}\left|a_{n}\right|<\infty\) and \(\sum_{n=0}^{\infty}b_{n}<\infty\), then \(\sum_{n=0}^\infty c_n<\infty\)

More specifically, if \(\sum_{n=0}^{\infty}\left|a_{n}\right|=A\) and \(\sum_{n=0}^\infty b_n = B,\) then: \(\sum_{n=0}^\infty c_n = A \cdot B.\)

Proof

Let \(A_n := \sum_{k=0}^{n} a_k\) and \(B_n := \sum_{k=0}^{n} b_k\), \(\mathcal{C}_{n}:=\sum_{k=0}^{n}c_{k}\)

Hypothesis: \(\lim_{n \to \infty} A_n = A\) and \(\lim_{n \to \infty} B_n = B\) Thesis: \(\lim_{n \to \infty} \mathcal{C}_n = A \cdot B\)

\(\mathcal{C}_{n}=\sum_{k=0}^{n}c_{k}=c_0+c_1+c_2+\dots+c_{n}\\=a_0b_0+\\(a_0b_1+a_1b_0)+\\(a_0b_2+a_1b_1+a_2b_0)+\\\dots+\\(a_0b_{n}+a_1b_{n-1}+\dots+a_{n}b_0)\\=a_0B_{n}+a_1B_{n-1}+\dots+a_{n}B_0\)

Now we need to use \(\lim_{n \to \infty} B_n = B\), thus we let \(B\) involved in this equation.

Let \(\beta_n = B - B_n = \sum_{k=n+1}^{\infty} b_k \Rightarrow B_n = B - \beta_n\)

\(\mathcal{C}_{n}=a_0(B-\beta_{n})+a_1(B-\beta_{n-1})+\dots+a_{n}(B-\beta_0)\)

\(= a_0 B + a_1 B + \dots + a_n B - (a_0 \beta_n + a_1 \beta_{n-1} + \dots + a_n \beta_0)\)

\(= (a_0 + a_1 + \dots + a_n) B - \gamma_n\)

\(=A_{n}B-\gamma_{n}\) where \(\gamma_n := (a_0 \beta_n + a_1 \beta_{n-1} + \dots + a_n \beta_0)\)

\(\Rightarrow\mathcal{C}_{n}=A_{n}B-\gamma_{n}\) \(\text{where } A = \lim_{n \to \infty} A_n\)

\(\Rightarrow\lim_{n\to\infty}\mathcal{C}_{n}=A_{}B-\lim_{n\to\infty}\gamma_{n}\)

We need to prove that \(\lim_{n\to\infty}\gamma_{n}=0\Rightarrow\lim_{n\to\infty}\left(a_0\beta_{n}+a_1\beta_{n-1}+\ldots+a_{n}\beta_0\right)=0\)

Which is \(\forall \varepsilon>0,\exists n_0,n>n_0:|\gamma_n|<\varepsilon\)

Thus we want \(|\gamma_{n}|=|a_0\beta_{n}+a_1\beta_{n-1}+\cdots+a_{n_0}\beta_{n-n_0}+a_{n_0+1}\beta_{n-n_0-1}+\cdots+a_{n}\beta_0|\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\leq|a_0||\beta_{n}|+\cdots+|a_{n_0}||\beta_{n-n_0}|+|a_{n_0+1}||\beta_{n-n_0-1}|+\cdots+|a_{n}||\beta_0|\)

Then since \(\left|a_0\right|,\left|a_1\right|,\ldots,\left|a_{n_0}\right|\) is finite, then we can bound it: \(|a_{n}|<M_{a}\quad\forall n\leq n_0\)

Also since \(\left|\beta_{n}\right|,\left|\beta_{n-1}\right|,\ldots,\left|\beta_{n-n_0}\right|\) is finite, then we can bound it: \(|\beta_{n}|<M_{\beta}\quad\forall n\in\left\lbrack n-n_0,n\right\rbrack\)

Then we have \(|a_0||\beta_{n}|+\cdots+|a_{n_0}||\beta_{n-n_0}|+|a_{n_0+1}||\beta_{n-n_0-1}|+\cdots+|a_{n}||\beta_0|\\\leq M_{a}M_{\beta}+\sum_{k=n_0+1}^{n}\left|a_{k}\right|\left|\beta_{n-k}\right|\)

Since \(\lim B_{n}=B\), then \(\lim_{n\to\infty}B-B_{n}=0\Rightarrow\forall\varepsilon_1>0,\exists n_1,n>n_1:\left|\beta\right|<\varepsilon_1\)

Then we have \(M_{a}M_{\beta}+\sum_{k=n_0+1}^{n}\left|a_{k}\right|\left|\beta_{n-k}\right|\leq M_{a}M_{\beta}+\varepsilon_1\cdot\sum_{k=1}^{\infty}\left|a_{k}\right|\leq M_{a}M_{\beta}+\varepsilon_1A\)

Also we need to take \(n>\max\{n_0,n_1\},\varepsilon_1=\frac{\varepsilon-M_{a}M_{\beta}}{A}\)

Finally \(|\gamma_n|\leq\varepsilon\)

Therefore \(\lim_{n \to \infty} \mathcal{C}_n = A \cdot B\)