Sum of inverse of two divergent sequences

We can see clearly that ${1\over a_n} - {1\over b_n} < {1\over b_{n-1}}- {1\over b_n}$

$ {1\over b_{n-1}}- {1\over b_n} $ is a telescopic sequence and since $\lim_{n\to \infty} {1\over b_n} = 0$ then $\sum {1\over b_{n-1}}- {1\over b_n} $ converges.

By comparaison of two positive series, $ \sum{1\over a_n}- {1\over b_n} $ converges too.

Let $c_n$ be the sequence $c_1=\frac{1}{a_1}$, $c_2=\frac{1}{b_1}$,$c_3=\frac{1}{a_2}$, $c_4=\frac{1}{b_2}$, etc.

Its terms are all positive, the sequence is monotone decreasing, and the sequence converges to $0$. Applying the Leibniz criterion, we reach the slightly stronger conclusion that the series

$$S=\sum(-1)^{n+1}c_n$$

converges. This implies the convergence of your desired series in particular (the partial sums of your series are a subsequence of the partial sums of $S$).

Let $c_n=\frac{1}{a_n}-\frac{1}{b_n}$. Then $c_n>0$ for all $n$, hence $$ 0<\sum_{n=k}^mc_n<\sum_{n=k}^m\Big(\frac{1}{a_n}-\frac{1}{a_{n+1}}\Big)=\frac{1}{a_k}-\frac{1}{a_{m+1}} $$ using the fact that the second sum telescopes.

Now since $a_n\to\infty$, for any $\varepsilon>0$ we can choose $N$ large enough that $a_n>\frac{1}{\varepsilon}$ for all $n\geq N$. Therefore $0<\sum_{n=k}^mc_n<2\varepsilon$ for all $m>k\geq N$, so $\sum_nc_n$ converges by the Cauchy criterion.