$\sum_1^\infty a_n$ converges iff $\sum_0^\infty 2^ka_{2^k}$ converges (Rudin)

Question

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?

Answer 1

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 $2^{k-1}$.

Answer 2

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.