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I must show that if a series $\sum_{n=1}^{\infty} a_n$ with positive terms converges, then the series $\sum_{n=1}^{\infty} A_n$, where $A_n = \sqrt{\sum_{k=n}^{\infty} a_k} - \ \sqrt{\sum_{k=n+1}^{\infty} a_k},$ also converges, and $a_n = o(A_n)$ as $n\to\infty$.

Here is my (rough) attempt:

Since $\sum_{n=1}^{\infty}a_n$ is positive and finite, $\sqrt{\sum_{k=n}^{\infty}a_k} $ is defined $\forall n \in \Bbb N$.

By the Cauchy criterion, for all $\frac{\varepsilon^2}{4} > 0,$ there is a $N \in \Bbb N$ such that $|\ a_n+\ ...\ +a_m \ | < \frac{\varepsilon^2}{4}$ for all $m \geq n > N$.

Then for $| \ A_n+\ ...\ +A_m | = \left|\sqrt{\sum_{k=n}^{\infty} a_k} - \ \sqrt{\sum_{k=m+1}^{\infty} a_k} \ \right| \leq \left|\sqrt{\sum_{k=n}^{\infty} a_k}\right| + \left|\sqrt{\sum_{k=m+1}^{\infty} a_k}\right| $=

$ \sqrt{ \left | \sum_{k=n}^{\infty} a_n \right| }$ + $\sqrt{ \left | \sum_{k=m+1}^{\infty} a_n \right| }$ $ < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon $, where $m \geq n > N$. Thus $A_n$ converges.

For $$\lim \limits_{n \to \infty} \frac{a_n}{ \left(\sqrt{\sum_{k=n}^{\infty} a_k} - \ \sqrt{\sum_{k=n+1}^{\infty} a_k}\right)} = \lim \limits_{n \to \infty} \frac{a_n \cdot \left( \sqrt{\sum_{k=n}^{\infty} a_k} + \ \sqrt{\sum_{k=n+1}^{\infty} a_k}\right)}{\left( \sqrt{\sum_{k=n}^{\infty} a_k} - \ \sqrt{\sum_{k=n+1}^{\infty} a_k}\right)\cdot\left( \sqrt{\sum_{k=n}^{\infty} a_k} + \ \sqrt{\sum_{k=n+1}^{\infty} a_k}\right)} = \lim \limits_{n \to \infty} \frac{a_n \cdot \left( \sqrt{\sum_{k=n}^{\infty} a_k} + \ \sqrt{\sum_{k=n+1}^{\infty} a_k}\right)}{ \left(\sum_{k=n}^{\infty} a_k - \ \sum_{k=n+1}^{\infty} a_k\right)} = \lim \limits_{n \to \infty} \frac{a_n \cdot \left( \sqrt{\sum_{k=n}^{\infty} a_k} + \ \sqrt{\sum_{k=n+1}^{\infty} a_k}\right)}{ a_n} = \lim \limits_{n \to \infty} \sqrt{\sum_{k=n}^{\infty} a_k} + \ \sqrt{\sum_{k=n+1}^{\infty} a_k} = 0 + 0 = 0 $$

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Fix some sequence $(b_k)_{k\geqslant1}$ and define $c_k=b_k-b_{k+1}$ for every $k\geqslant1$. Then, for every $n\geqslant0$, $$ \sum_{k=1}^nc_k=b_1-b_{n+1}. $$ If the sequence $(b_k)$ converges, this implies that the series $\sum\limits_kc_k$ converges and that its sum is $b_1-b_\infty$, where $b_\infty=\lim\limits_{k\to\infty}b_k$.

Applying this to $$ b_k=\sqrt{\sum_{i=k}^\infty a_i}, $$ one sees that the series $\sum\limits_kA_k$ converges and that its sum is $$ \sum_{k=1}^\infty A_k=\sqrt{\sum_{i=1}^\infty a_i}. $$ To prove that $a_n\ll A_n$, note that $$ a_n=A_n\cdot\left(b_n+b_{n+1}\right),\qquad b_n+b_{n+1}\to0. $$

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  • $\begingroup$ Your approach is obviously much simpler than my convoluted method, but I would like to know whether the first part of mine was correct. $\endgroup$ – user110503 Apr 5 '14 at 20:17
  • $\begingroup$ Before "Thus An converges", indeed what you do show that the series $\sum\limits_nA_n$ converges. $\endgroup$ – Did Apr 5 '14 at 20:26

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