A analytic representation of q- rational series

Question

Using Mathematica, we can find $$\sum\limits_{n = 1}^\infty {\frac{{{{\left( {1 - q} \right)}^2}{q^n}}}{{\left( {1 - {q^n}} \right)\left( {1 - {q^{n + 1}}} \right)}}} = q,\;q \in \left( {0,1} \right).$$ On the other hand, we can easily seen that $$\sum\limits_{n = 1}^\infty {\frac{{{{\left( {1 - q} \right)}^2}{q^n}}}{{\left( {1 - {q^n}} \right)\left( {1 - {q^{n + 1}}} \right)}}} = \sum\limits_{n = 1}^\infty {\left\{ {\frac{{1 - q}}{{1 - {q^n}}} - \frac{{1 - q}}{{1 - {q^{n + 1}}}}} \right\}} = 1.$$ Why?

Answer 1

We have: $$\sum_{n=1}^{N}\frac{(1-q)^2 q^n}{(1-q^n)(1-q^{n+1})}=\sum_{n=1}^{N}\left(\frac{1-q}{1-q^n}-\frac{1-q}{1-q^{n+1}}\right) = 1-\frac{1-q}{1-q^{N+1}}$$ hence as $N\to +\infty$ the RHS of the previous line tends to $1-(1-q)=\color{red}{q}$.

Answer 2

Check the telescoping series again. We have

$$\sum_{n=1}^N \left(\frac{1-q}{1-q^n}-\frac{1-q}{1-q^{n+1}}\right)=\frac{1-q}{1-q}-\frac{1-q}{1-q^{N+1}}$$

That second term above does not tend to $0$ as $N\to\infty$, it tends to $\frac{1-q}{1-q^\infty}=1-q$.