Proof of sandwich/squeeze theorem for series.

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.

We have that

$$a_{n}<b_{n}<c_{n}\iff 0<b_{n}-a_{n}<c_{n}-a_{n}$$

then the positive series

$$\sum b_{n}-a_{n}$$

converges for comparison test and therefore

$$\sum b_{n}=\sum (a_{n}-(b_{n}-a_{n}))=\sum a_{n}-\sum (b_{n}-a_{n})$$

converges since it is the sum of two convergent series.