# Improper Integral vs. Infinite Series

$\displaystyle \int_{2}^{\infty}{\frac{1}{x^{2} - 1}dx} =\boxed{ \ln {3}} = 1.09$

$\displaystyle \sum_{n=2}^{\infty}{\frac{1}{n^{2} - 1}} = \boxed{\frac{3 } {2 }} = 1.5$

To test if the infinite series $\displaystyle \sum_{n=2}^{\infty}{\frac{1}{n^{2} - 1}}$ converges,
provided the function $\displaystyle f(n) = {\frac{1}{n^{2} - 1}}$
in the interval [2, ∞[ is:

• positive &
• decreasing function

you can convert it to $\displaystyle \int_{2}^{\infty}{\frac{1}{x^{2} - 1}dx}$
(although they converge to different values.)