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Given three series $A_n \leq B_n \leq C_n$, I should prove that $B_n$ is convergent if the series of the partial sums of $A_n$ and $C_n$ are convergent.

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@Mohamad: You're mixing up sequences and series. Usually one is interested in the sequence of partial sums of a sequence, or the series associated to a sequence (the series whose terms are precisely that sequence); and $A_n$, $B_n$, and $C_n$ are themselves sequences, not series. – Zev Chonoles Jun 24 '12 at 19:30
See this question – Julián Aguirre Jun 24 '12 at 21:13

Have you tried using squeeze theorem here. It is quite useful in calculating limits of functions. See it might work for series as well because you are using the concept of convergence of the series of partial sum of $A_n$ and $C_n$.

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To prove this, the hypothesis must be strengthened very slightly. We will assume not only that the partial sums are bounded, but also that the partial sums of $A_n$ and $C_n$ approach a limit, which we will denote $\sum A_n$ and $\sum C_n$, respectively.

By the first inequality, $C_n-A_n\ge B_n-A_n \ge 0$.

$\sum B_n-A_n$ is monotonic (since each successive term is positive) and is bounded above by $\sum C_n-A_n$. A monotonic sequence bounded above converges to a limit$^1$. Therefore, $\sum_{n=0}^{\infty} B_n - A_n$ exists and $\sum B_n$ converges.

$^1$This sequence is monotonic increasing, so the limit of the sequence is the least upper bound of the sequence itself.

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??? A sequence can be bounded above and below and still not converge. I suspect that Mohamad is leaving out some important details of the problem. – Robert Israel Jun 24 '12 at 19:35
I need to edit it still...sorry. – Andrew Salmon Jun 24 '12 at 19:37
How do you know that $\sum C_n-A_n$ converges? – Jose27 Jun 24 '12 at 20:24
Because $\sum C_n - A_n = \sum C_n - \sum A_n$, and $\sum C_n$ and $\sum A_n$ both converge. – Andrew Salmon Jun 24 '12 at 20:27

That need not to be true. Take an increasing sequence $A_n$ that converges to $-1$ and a decreasing sequence $C_n$ that converges to $1$. Then take $B_n = (-1)^n$. Clearly the relation you state holds, but $\lim B_n$ does not exist. What I think you're looking for is the so called squeeze theorem:


Let $a_n < b_n < c_n$ for all $n \in \Bbb N$. Then if $\lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} c_n=\mu \in \Bbb R$ we also have $\lim\limits_{n \to \infty} b_n = \mu $.

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