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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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    See [this question](http://math.stackexchange.com/questions/157853/proof-of-a-test-for-series/157862#157862)2012-06-24

3 Answers 3

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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:

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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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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    Because $\sum C_n - A_n = \sum C_n - \sum A_n$, and $\sum C_n$ and $\sum A_n$ both converge.2012-06-24