# For $(a_n),(b_n),(c_n)$, with $a_n\leq b_n\leq c_n$, convergence of $\sum\limits_{n=0}^\infty b_n$

Let $$(a_n),(b_n),(c_n)$$ be sequences of real numbers with the property that for each $$n\in\mathbb{N}$$, $$a_n\leq b_n\leq c_n$$. Suppose both series, $$\sum\limits_{n=1}^\infty a_n$$ and $$\sum\limits_{n=0}^\infty c_n$$ converge. Then prove that $$\sum\limits_{n=0}^\infty b_n$$ also converge.

I read the similar post in here. But couldn't relate it to my question. Appreciate any help.

• Sorry. I edited the question – DD90 Jan 7 at 23:13
• Have you tried to use the Cauchy sequence criterion? – Mindlack Jan 7 at 23:14
• @Mindlack gives a good hint. If the sum of $a_n$'s and $b_n's$ converged to the same thing, there would be an obvious candidate for the sum of the $b_n's$. However, you're not given that, so the sum of the $b_n's$ can kind of do whatever it wants as long as it stays between the bounds. So it is in this case, that you have no idea what the sum would converge to, that the Cauchy cirterion is taylored for. – Ovi Jan 7 at 23:17

This is an immediate corollary to the comparison test. Consider $$0 \le b_n - a_n \le c_n - a_n.$$ Then, $$\sum (c_n - a_n)$$ is a convergent, positive series, hence so is $$\sum (b_n - a_n)$$ (the partial sums of $$\sum (b_n - a_n)$$ form a monotone sequence, bounded above by $$\sum (c_n - a_n)$$; apply monotone convergence theorem). Since $$\sum a_n$$ is convergent, then so is $$\sum b_n = \sum (b_n - a_n) + \sum a_n$$.

• So here does $b_n$ means the partial sums of the $\sum b_n$? – DD90 Jan 7 at 23:19
• No, just the individual terms. The individual terms of the series $\sum b_n - a_n$ are positive and dominated by the individual terms of the convergent series $\sum c_n - a_n$. – Theo Bendit Jan 7 at 23:21
• Yes, that's right. It's proven (as I alluded to in the answer) with the monotone convergence theorem, which requires positive terms in order to make the sequence of partial sums monotone. Essentially, your question is the non-positive version; it's ok to have negative terms, provided the terms are bounded below by a convergent (possibly negative) series. – Theo Bendit Jan 7 at 23:28
• Thank you! Appreciate your help – DD90 Jan 7 at 23:30
• "$\sum c_n - a_n$" Better to write $\sum (c_n - a_n)$ – zhw. Jan 7 at 23:34