# Proof of sandwich/squeeze theorem for series.

I am interested in proving a theorem, which I suppose one may call a sandwich or squeeze theorem for series.

Suppose we have three series: $\sum^{\infty}_{n=1}a_{n}$, $\sum^{\infty}_{n=1}b_{n}$ and $\sum^{\infty}_{n=1}c_{n}$. We know that $\sum^{\infty}_{n=1}a_{n}$ and $\sum^{\infty}_{n=1}c_{n}$ converge; furthermore, let us assume that for all $n\in\mathbb{N}$, the following inequality holds: $a_{n}<b_{n}<c_{n}$. Then, the series $\sum^{\infty}_{n=1}b_{n}$ will also converge.

The only way that I can think of to approach the proof of the above would be via the Cauchy criterion, i.e. showing that

$$\forall_{\varepsilon>0}\,\exists_{n_{\varepsilon}}\,\forall_{m>k>n_{\varepsilon}}\quad|b_{k+1}+...+b_{m}|<\varepsilon.$$

As I understand, in order to show that, we would have to somehow bound $\left|\sum^{m}_{n=k+1}b_{n}\right|$ by $\left|\sum^{m}_{n=k+1}c_{n}\right|$ and/or $\left|\sum^{m}_{n=k+1}a_{n}\right|$. If we assume nonnegativity of the terms of $\sum{}b_{n}$ and $\sum{}c_{n}$, the task becomes trivial. However, without this assumption, I am having problems with finding the right bound.

I would be thankful for some hints on how this could be done, or possibly advice on a different approach to the proof. Thank you in advance.

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Does series $a_n, c_n$ converge to the same number? – SundayCat Nov 2 '13 at 19:03
Nope, they don't have to. – Johnny Westerling Nov 2 '13 at 20:11
I think you should formalize your assumption here, making it more clear. If they don't converge to the same number. What are you trying to prove? Prove the series $b_n$ also converge? – SundayCat Nov 2 '13 at 20:14
I believe I formulated everything quite clearly... we have three series, we know that $\sum{}a_{n}$ and $\sum{}c_{n}$ converge, and we have the inequality $a_{n}<b_{n}<c_{n}$ for all $n\in\mathbb{N}$. We are trying to prove that under such conditions, $\sum{}b_{n}$ converges. Could please say explicitly what is unclear? – Johnny Westerling Nov 2 '13 at 20:21
I am just a little bit confused here. Since the inequality $b_n \lt c_n$, it implies that $\sum_{i=1}^{n}b_i \lt \sum_{i=1}^{n}c_i$. Also because $\sum_{i=1}^{\infty}c_i$ is convergent, then $\sum_{i=1}^{\infty}b_i$ is also convergent by the comparison test. We did not need the condition $a_n \lt b_n$ here. – SundayCat Nov 2 '13 at 20:27

Let $\varepsilon > 0$ be given. By the convergence of the series $\sum a_n$ resp. $\sum c_n$, there is an $N(\varepsilon)$ such that for all $N(\varepsilon) \leqslant k < m$ we have

$$\left\lvert\sum_{n=k+1}^m a_n \right\rvert < \varepsilon\land \left\lvert\sum_{n=k+1}^m c_n \right\rvert < \varepsilon.\tag{1}$$

By the inequalities $a_n \leqslant b_n \leqslant c_n$, we have

$$\sum_{n=k+1}^m a_n \leqslant \sum_{n=k+1}^m b_n \leqslant \sum_{n=k+1}^m c_n.$$

Now $(1)$ implies

$$-\varepsilon < \sum_{n=k+1}^m b_n < \varepsilon$$

for $N(\varepsilon) \leqslant k < m$.

Since $\varepsilon$ was arbitrary, the sequence of partial sums is a Cauchy sequence, and $\sum b_n$ converges.

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OK, I see how it works. Thank you. – Johnny Westerling Nov 2 '13 at 20:37