Sum of inverse of two divergent sequences Let $a_n, b_n$ be two real sequences for wich $0<a_n<b_n<a_{n+1}$. Let $\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=\infty$
I need to prove that $\sum{\frac{1}{a_n}-\frac{1}{b_n}}$ is convergent.
It's been a long time since I did some real analysis, so I probably forgot some basic thing to prove this, because I don't think that'll be very hard.
PS: English isn't my main language, forgive me some errors.
 A: 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 $S$).
A: 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.
A: 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.
