I have a question with "Intoduction to analysis, fourth edition" by Wade, theorem 6.40.

Question

I have a question with the theorem 6.40 ii) from the book "Introduction to Analysis, fourth edition", which is written by Wade. Theorem 6.40

I think if $k>N$, $|a_{N + 1}| \leq |a_ N| \times x$ cannot be established, so I think it should be $k \geq N$.

So I want to know that am I right.

If I'm wrong, please explain the reason to me.

Thanks!

Answer 1

Suppose $x\in (0,1)$ and $s= \sum_{k=0}^\infty |a_k|$ for reals $a_k$. Further, for $k> N$, $|\frac{a_{k+1}}{a_k}|<x$.

Prove $0\le s - \sum_{k=0}^n|a_k|\le \frac{|a_N|x^{n-N+1}}{1-x} \forall n\ge N$.

I think $k\ge N$ works in part because it includes $k>N$. I think he defined it that way to save you a step.

By hypothesis, $|\frac{a_{N+1}}{a_N}|<x$. So$ |a_{N+1}|<x|a_N|<|a_N|$.

We also have by similar arguments $|a_{N+2}|<x|a_{N+1}|<|a_{N+1}|$.

Combined, these imply $|a_{N+2}|<x^2|a_N|$, and more generally, $|a_{N+k}|<x^k|a_N|$.

Thus it follows $|a_N|+|a_{N+1}|+|a_{N+2}|+...\le \frac{|a_N|}{1-x}$

Similary, by hypothesis, this holds for any $n>N$, so$ |a_n|+|a_{n+1}|+|a_{n+2}|+...\le \frac{|a_n|}{1-x}$

We also have by above arguments that $|a_n|<x^{n-N}|a_N|$

So $s\le\sum_{k=0}^{n-1}|a_k|+\frac{|a_N|x^{n-N}}{1-x}$

Use $n+1$ inplace of n instead, and subtract the sigma expression from both sides, we get:

$s-\sum_{k=0}^n|a_k|\le \frac{|a_N|x^{n-N+1}}{1-x}$

Answer 2

I Think I'm right. For example, if
$$a_k = \begin{cases} 0 & \text{if k = 1}\\ (1/3)^k & \text{otherwise} \end{cases},$$
$\sum_{k=1}^{\infty} {a_k}=1/6$(absolute converge). And if $k>N=1$, $$\frac{|a_{k+1}|}{|a_{k}|} \leq \frac{1}{3} .$$
Meanwhile, $s-\sum_{k=1}^{1} {a_k} = 1/6 - 0 = 1/6$ and
$$\frac{|a_N|x^{n-N+1}}{1-x} = 0$$
so,$$s-\sum_{k=1}^{1} {a_k} > \frac{|a_N|x^{n-N+1}}{1-x} .$$
So, I think it should be fixed to $k\geq N$.