proving that a sum is absolutely convergent iff another sum is absolutely convergent

Question

I'm having trouble with the following question:

Prove that $\sum_{n=1}^\infty\left(\frac{1}{a_{n+1}}-\frac{1}{a_n}\right)$ is absolutely converges if and only if $\sum_{n=1}^\infty(a_{n+1}-a_n)$ absolutely converges, where $(a_n)$ is a series whose elements are non-zero and $a_n \to a \neq 0$.

What I tried to do:

I subtracted the fractures of $\sum_{n=1}^\infty\left(\frac{1}{a_{n+1}}-\frac{1}{a_n}\right)$ and then I've taken the other sum, $\sum_{n=1}^\infty(a_{n+1}-a_n)$ and multiplied it by $\frac{(a_{n+1}-a_n)}{(a_{n+1}-a_n)}$ to find a relation between the two. Then in order to show that both are absolutely converging, I tried to show that their absolute values must also converge as a rule, so basically from what I understand each elements of both sums should be positive for it to pass the absolute convergence test.

However I got lost there. How to show correctly that the first sum $\sum_{n=1}^\infty\left(\frac{1}{a_{n+1}}-\frac{1}{a_n}\right)$ is absolutely convergent iff the second sum $\sum_{n=1}^\infty(a_{n+1}-a_n)$ is also absolutely convergent?

Answer 1

Use the limits comparison theorem:

$$\lim_{n\to\infty}\frac{\left|\frac1{a_{n+1}}-\frac1{a_n}\right|}{|a_{n+1}-a_n|}=\lim_{n\to\infty}\frac1{|a_{n+1}a_n|}=\frac1{a^2}>0$$

and thus both your series converge or diverge together.

Answer 2

Alternately, we could observe that $$ \sum_{n=1}^\infty(-a_na_{n+1})\left(\frac1{a_{n+1}}-\frac1{a_n}\right)=\sum_{n=1}^\infty\left(a_{n+1}-a_n\right), $$ and make use of the fact that if $\sum b_n$ converges absolutely and $\{c_n\}$ is a bounded sequence, then $\sum b_nc_n$ also converges absolutely.