Assume $\sum \frac{1}{a_n} < \infty$, How about $\sum \frac{1}{a_n - 1}$? Problem :
Given sequence $\left\{a_n\right\}$ s. t. $\forall n \in \mathbb{N},a_n(a_n-1)\neq 0$.
If $\displaystyle\sum_{n=1}^\infty\frac{1}{a_n}$ converge, then how about $\displaystyle\sum_{n=1}^\infty\frac{1}{a_n-1}$?

Rewrite $$\frac{1}{a_n-1} = \frac{1}{a_n}\times \frac{1}{1-\frac{1}{a_n}}$$
Since $\displaystyle\sum_{n=1}^\infty\frac{1}{a_n}$ converge, we know that $\displaystyle\frac{1}{a_n}\to 0$.
From $\displaystyle\frac{1}{1-x}\approx1+x$,   $\quad\displaystyle\frac{1}{a_n-1} = \frac{1}{a_n}\times \frac{1}{1-\frac{1}{a_n}}\approx\frac{1}{a_n}\times\left(1+\frac{1}{a_n}\right)=\frac{1}{a_n}+\frac{1}{a_n^2}$
And we know both $\displaystyle\sum\frac{1}{a_n},\quad\sum\frac{1}{a_n^2}$ converge, finally $\displaystyle\sum\frac{1}{a_n-1}$ also converge.

Is this proof valid? if not, I wonder to know the counterexample for this one.
Thank you.

For the same example given by Lorago (assuming $n>1$), notice that $\dfrac{1}{a_n(a_n-1)}$ would be positive for sufficiently large $n$ (in fact positive for every $n>1$). Specifically, $$\dfrac{1}{a_n(a_n-1)} = \dfrac{1}{n+(-1)^{n+1}\sqrt{n}}.$$
Thus $\displaystyle\sum \dfrac{1}{a_n(a_n-1)} $(considering from the stage when it turns positive) is divergent by limit comparison test (compare it with $\displaystyle\sum \dfrac{1}{n})$.
Notice also that
$$\sum\left( \dfrac{1}{a_n-1} -\dfrac{1}{a_n}\right)= \sum \dfrac{1}{a_n(a_n-1)}.$$
Hence $\displaystyle\sum\dfrac{1}{a_n-1} $ must be divergent.
