# if $\sum_{n}^{\infty}a_n$ converges, then $\sqrt{\sum_{k=n}^{\infty}a_k} -\sqrt{\sum_{k=n+1}^{\infty}a_k}$ also converges, and consequence

Let $a_1, a_2, ..., a_n$ be positive sequence, then if $\sum_{n}^{\infty}a_n$ converges, prove that:

$$A_n = \sqrt{\sum_{k=n}^{\infty}a_k} -\sqrt{\sum_{k=n+1}^{\infty}a_k}$$

$\sum_{n}^{\infty}A_n$ also converges.

$$\sum_{k=1}^{\infty}a_k = S$$

So $$\sum_{n}^{\infty}A_n = \sqrt{S} - \sqrt{S-a_1}+\sqrt{S-a_1} - \sqrt{S-a_1 - a_2} ... = \sqrt{S} - \sqrt{S - S_k}$$

So it is convergent.

1) My textbook says that $a_n = o(A_n)$. I can't see that from my equations.

2) Also, I have a question about n approaching infinity. Can we say that $S_k$ at this base will be equal to $S$. What confuses me - k approaches n, but n approaches infinity, it is right to consider that in infinity k will be equal to n?

There is similar equation for divergent series.

Let $a_1, a_2, ..., a_n$ be positive sequence, then if $\sum_{n}^{\infty}a_n$ diverges, then:

$$A_n = \sqrt{\sum_{k=1}^{n}a_k} -\sqrt{\sum_{k=1}^{n-1}a_k}$$

$\sum_{n}^{\infty}A_n$ also diverges.

3) My textbook says that all this implies that

There can not exist reference series to compare with to test on convergence.

I've been thinking for a while and I can't see why this conclusion comes from this exact derivations.

Let $S_n=\sum_{k=n}^\infty a_k$. Then $\lim_{n\to\infty}S_n=0$. Now $$A_n=\sqrt{S_n}-\sqrt{S_{n+1}}=\frac{S_n-S_{n+1}}{\sqrt{S_n}+\sqrt{S_{n+1}}}=\frac{a_n}{\sqrt{S_n}+\sqrt{S_{n+1}}},$$ and $$\lim_{n\to\infty}\frac{a_n}{A_n}=\lim_{n\to\infty}(\sqrt{S_n}+\sqrt{S_{n+1}})=0\implies a_n=o(A_n).$$ This shows that there is no lower bound on the rate of vconvergence towards $0$ at $\infty$ of the terms of a convergent series; there will always be another convergent series whose terms converge to $0$ more slowly.