Prove that $\sum_{n=1}^\infty |\frac{1}{a_{n+1}} - \frac{1}{a_n}| \space converge \Longleftrightarrow \sum_{n=1}^\infty |a_{n+1} - a_n|$ converge

Question

let $(a_n)$ be a sequence, $\forall n, a_n \ne 0$, and $a_n \to a \ne 0$ so assuming $a_n \ne a_{n+1}$ I have already proven that: $$\lim_{n\to \infty} \frac{|a_{n+1} - a_n|}{|\frac{1}{a_{n+1}} - \frac{1}{a_n}|}=\lim_{n\to \infty} \frac{|a_{n+1} - a_n|}{|\frac{a_n - a_{n+1}}{a_{n+1}\cdot a_n}|}=\lim_{n\to \infty} |a_{n+1}\cdot a_n| = a^2 >0$$ and from the limit comparison test I get that $$\sum_{n=1}^\infty (\frac{1}{a_{n+1}} - \frac{1}{a_n}) \space absolutely \space converge \Longleftrightarrow \sum_{n=1}^\infty (a_{n+1} - a_n) \space absolutely \space converge$$ but I don't know what to do if the are indexes, or a sequence of indexes for which $a_{n+1} = a_n \Longrightarrow a_{n+1} - a_n=0$ in which case I can't use the LCT

Answer 1

The statement is not true I think. For example, take $a_n=1/n$.

Then $\sum_{n=1}^\infty |\frac{1}{a_{n+1}} - \frac{1}{a_n}|=\sum_{n=1}^\infty 1$, which is diverging, while $\sum_{n=1}^\infty |a_{n+1} - a_n|=\sum_{n=1}^\infty(\frac{1}{n}-\frac{1}{n+1})$, which is clearly converges to 1.

I miss the assumption $\lim_{n}a_n \neq 0$. So for your question, with the indexes for which $a_{n+1}=a_n$. Then you can ignore the term $\mid a_n-a_{n+1}\mid$,, delete $a_{n+1}$ and just consider the new sequence $a_n$ with properties $a_n \neq a_{n+1}$ without changing anything.