# Carry convergence from one series to another

Suppose I have

(i) $\alpha_n \geq 0$, $\beta_n \geq 0$ such that

(ii) $\sum_{n=1}^\infty \alpha_n = \sum_{n=1}^\infty \beta_n = 1$, and

(iii) $\sum_{n=1}^\infty n\alpha_n < \infty$. Intuitively, if

(iv) $\forall N \in \mathbb{N} :\sum_{n=1}^N \alpha_n \leq \sum_{n=1}^N \beta_n$

then I should also get $\sum_{n=1}^\infty n\beta_n < \infty$.

(Sorry, there was a typo in the original equation (iv): the direction of the inequality was reversed). Here is my proposed solution, but I feel queasy about some steps so please point out any mistakes.

$\sum_{n=1}^\infty n \alpha_n = \sum_{n=1}^\infty \sum_{m=n}^\infty \alpha_n = \sum_{n=1}^\infty \left(1 - \sum_{m=1}^{n-1} \alpha_n \right)$

$\geq \sum_{n=1}^\infty \left(1 - \sum_{m=1}^{n-1} \beta_n \right) = \sum_{n=1}^\infty \sum_{m=n}^\infty \beta_n = \sum_{n=1}^\infty n \beta_n$

• actually I have a solution, but I don't know if it is correct; I will edit to post to add it, so if anyone could point out correctness or mistakes, I would appreciate it. – mitchus Jan 21 '13 at 17:51
• Your outline is fine. – Andrés E. Caicedo Jan 21 '13 at 18:05
• @AndresCaicedo Thanks. I would like to better understand the basic concepts to follow when manipulating series, in order to be more comfortable in the future -- is there a textbook which you would recommend? – mitchus Jan 23 '13 at 9:23
• @AndresCaicedo : actually, looking back at this, I have trouble justifying the first step: $\sum_{n=1}^\infty n \alpha_n = \sum_{n=1}^\infty \sum_{m=n}^\infty \alpha_n$. – mitchus Jul 8 '13 at 9:04
• I guess that should be $\sum_{n=1}^\infty n \alpha_n = \sum_{n=1}^\infty \sum_{m=n}^\infty \alpha_m$. – mitchus Jul 8 '13 at 9:09