test the convergence of an infinite series How to prove that the $\displaystyle \sum_{n=1}^{+\infty} (1-e^{(-1/n^2)})$ series is convergent? I can not find a number to use the comparison test!

Hints: 
$$  e^{-x } = \lim_{n} \left({}1 - \dfrac {x} {n}\right)^n \ge  \left({}1 - \dfrac {x} {n}\right)^n  \ge 1 - n \cdot  \dfrac x n = 1 - x  \;\;\;\; \forall x \gt 0 \;\;\; $$
This is due to Bernoulli's Inequality and due to the fact that the sequence $\lim_{n} \left({}1 - \dfrac {x} {n}\right)^n    $ is monotone increasing. Hence, 
$$  e^{-\frac{1}{n^2}} \ge \left({}1 - \dfrac{1}{n^2} \cdot \dfrac {1} {n}\right)^n \ge 1 - \dfrac{1}{n^2} $$
Now you can apply the Comparison Test as you wish to do. 
