Let $(a_n)_{n∈N}$ be a sequence of positive numbers which tends to zero but such that $\sum_{n=1}^\infty a_n$ diverges. Let $(A_n)_{n∈N}$ be the sequence of partial sums $A_n=\sum_{k=1}^n a_k$,and let $b_{n+1}=\sqrt{A_{n+1}}−\sqrt{A_{n}}$. Show that $\lim_{n→∞}\frac{b_n}{a_n}= 0$ but that $\sum b_n$ is still divergent.
Question 1:
I was able to answer the first part, since
$$\frac{b_{n+1}}{a_{n+1}} = \frac{\sqrt{A_{n+1}}−\sqrt{A_{n}}}{a_{n+1}} = \frac{A_{n+1}-A_n}{a_{n+1}(\sqrt{A_{n+1}}+\sqrt{A_{n}})} = \frac{1}{\sqrt{A_{n+1}}+\sqrt{A_{n}}}$$ which converges to $0$ since $\sqrt{A_n}$ tends to infinity.
The second part, I have no idea about. Apparently $$\sum b_k = \sqrt{A_n}$$ Why is this true?
Question 2: It is also said that
In this sense there is no ‘smallest’ divergent series, and one can similarly show that there is no ‘largest’ convergent one.
I don't see why this follows. What about our findings conveys these ideas?
We have$$\sum_{k=1}^{n}b_{k+1}{=\sum_{k=1}^{n}\sqrt{A_{k+1}}-\sqrt{A_k}\\=\sqrt{A_{2}}-\sqrt{A_1}\\+\sqrt{A_{3}}-\sqrt{2}\\+\sqrt{A_{4}}-\sqrt{A_3}\\+\cdots\\+\sqrt{A_{n+1}}-\sqrt{A_n}}\\=\sqrt{A_{n+1}}-\sqrt{A_1}$$according to telescoping rule (https://en.wikipedia.org/wiki/Telescoping_series), $\sum_{k=1}^{n}b_{k+1}$ diverges since $\sqrt{A_{n+1}}-\sqrt{A_1}$ diverges.