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.
Best Answer
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.