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.

  • $\begingroup$ Oh I see... Thanks! $\endgroup$ – Joe Half Face Oct 14 '16 at 10:01

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.