Let $\{f_n\}$,$\{g_n\}$, and $\{h_n\}$ be sequences of functions on $\Bbb R$ that $f_n\le g_n \le h_n$ on $\Bbb R$ for all $n$. If $\sum f_n$ and $\sum h_n$ converges, then $\sum g_n$ converges.
This is what I think how to prove this statement. Given $x\in \Bbb R$, then $\sum f_n(x)$ and $\sum h_n(x)$ are Cauchy seqeunce in $\Bbb R$. For $\varepsilon>0$, there is $N\in \Bbb N$ such that $\forall n>m>N$, we have $$\left| \sum _{k=m+1}^nf_k(x) \right|<\varepsilon$$ and $$\left| \sum _{k=m+1}^nh_k(x) \right|<\varepsilon$$
Hence, $\forall n>m>N$, we have $$\left| \sum _{k=m+1}^ng_k(x) \right|=\left| \sum _{k=m+1}^n[g_k(x)-f_k(x)+f_k(x)] \right|\le \left| \sum _{k=m+1}^n[g_k(x)-f_k(x)] \right|+\left| \sum _{k=m+1}^nf_k(x) \right|\le \varepsilon+\left| \sum _{k=m+1}^n[h_k(x)-f_k(x)] \right|\le 3\varepsilon$$
Since $x\in \Bbb R $ is arbitrary, $\sum g_n$ converges.