2
$\begingroup$

enter image description here

enter image description here

enter image description here

Questions:

1-In the second part of the proof, why is $$a_1+a_2+(a_3+a_4)+...+(a_{2^{k-1}+1}+...+a_{2^k}) \ge \frac{1}{2} a_1+a_2+2a_4+...+2^{k-1}a_{2^k}$$ true?

2-Rudin says "The striking feature of the [that] theorem is that a rather 'thin' subsequence of ${a_n}$ determines the convergence or the divergence of $\sum a_n$". May someone explain this with an example?

$\endgroup$
  • $\begingroup$ Googling ‘Cauchy condensation test’ might help. $\endgroup$ – Jonatan B. Bastos Jan 21 '18 at 20:31
2
$\begingroup$

Answer to question 1: Observe that since $$a_1 \geq a_2 \geq a_3 \cdots \geq 0,$$we have $$a_3 + a_4 \geq a_4 + a_4,$$ and similarly

$ \underbrace{(a_{2^{k-1}+1}+...+a_{2^k})}_{\text{$2^{k-1}$ terms}} $ $\geq 2^{k-1}a_{2^k}$ where $a_{2^{k-1}+i} \geq a_{2^k}$ for each $i$ between 1 and $2^{k-1}$.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

For part 1 note that

$$a_1+a_2+(a_3+a_4)+...+(a_{2^{k-1}+1}+...+a_{2^k}) \ge \frac{1}{2} a_1+a_2+2a_4+...+2^{k-1}a_{2^k}$$

is true because each term in the LHS $\ge$to the corresponding term in the RHS

EG $$(a_3+a_4)\ge 2a_4$$

since $a_n$ is decreasing.

Note that the theorem is known as Cauchy condensation test which is useful to prove the convergence of $\sum \frac1{n^a}$.

For part 2 I think he is referring to the fact that with this criteria you can show convergence or divergence by considering only the $a_{2^k}$ terms.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.