Exercise 2.4.10

Question

[Infinite Products]
  A close relative of infinite series is the infinite product
  $$
  \prod_{n=1}^{\infty} b_{n}=b_{1} b_{2} b_{3} \cdots
  $$
  which is understood in terms of its sequence of partial products
  $$
  p_{m}=\prod_{n=1}^{m} b_{n}=b_{1} b_{2} b_{3} \cdots b_{m}
  $$

Consider the special class of infinite products of the form
  $$
  \prod_{n=1}^{\infty}\left(1+a_{n}ight)=\left(1+a_{1}ight)\left(1+a_{2}ight)\left(1+a_{3}ight) \cdots, \quad 	ext { where } a_{n} \geq 0
  $$
  \begin{enumerate}[label=(\alph*)] 
  \item Find an explicit formula for the sequence of partial products in the case where $a_{n}=1 / n$ and decide whether the sequence converges. Write out the first few terms in the sequence of partial products in the case where $a_{n}=1 / n^{2}$ and make a conjecture about the convergence of this sequence.
  \item Show, in general, that the sequence of partial products converges if and only if $\sum_{n=1}^{\infty} a_{n}$ converges. (The inequality $1+x \leq 3^{x}$ for positive $x$ will be useful in one direction.)
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item This is a telescoping product, most of the terms cancel
    $$
    p_m = \prod_{n=1}^m (1 + 1/n) = \prod_{n=1}^m \frac{n+1}{n} = \frac 21 \cdot \frac 32 \cdot \frac 42 \cdots \frac {m+1}{m} = m+1
    $$
    therefore $(p_m)$ diverges.

In the cast $a_n = 1/n^2$ we get
    $$
    \prod_{n=1}^\infty (1+ 1/n^2) = \prod_{n=1}^\infty \frac{1 + n^2}{n^2} = \frac 21 \cdot \frac 54 \cdot \frac{10}{9} \cdots
    $$ 
    The growth seems slower, I conjecture it converges now.
  \item Using the inequality suggested we have $1 + a_n \le 3^{a_n}$ letting $s_m = a_1 + \dots + a_m$ we get
    $$
    p_m = (1+a_1)\cdots(1+a_m) \le 3^{a_1}3^{a_2}\cdots 3^{a_m} = 3^{s_m}
    $$
    Now if $s_m$ converges it is bounded by some $M$ meaning $p_m$ is bounded by $3^M$. and because $a_n \ge 0$ the partial products $p_m$ are increasing, so they converge by the MCT. This shows $s_m$ converging implies $p_m$ converges.

For the other direction suppose $p_m 	o p$. Distributing inside the products gives $p_2 = a_1 + a_2 + 1 + a_1a_2 > s_2$ and in general $p_m > s_m$ implying that if $p_m$ is bounded then $s_n$ is bounded aswell. This completes the proof.

extbf{Summary}: Convergence is if and only if because $s_m \le p_m \le 3^{s_m}$.

(By the way the inequality $1 + x \le 3^x$ can be derived from $\log(1 + x) \le x$ implying $1 + x \le e^x$, I assume abbott rounded up to $3$.)
   \end{enumerate}

Exercise 2.4.10

Answers

Comments

Exercise 2.4.10

Answers

Comments

Add answer