First note that $$\lim_{x \to \tfrac{\pi}{2}^-} \sin(x) = 1$$ and $$\lim_{x \to \tfrac{\pi}{2}^-} \cos(x) = 0.$$
therefore $$\lim_{x \to \tfrac{\pi}{2}^-} \frac{\sin(x)}{\cos(x)} = + \infty$$ since we have (by the definition of limit):
$$\forall \varepsilon_1, \exists \delta_1, \forall x < \...In mathematics, the root test is a criterion for the convergence (a convergence test) of an infinite series. It depends on the quantity
:\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|},
where a_n are the terms of the series, and states that the series converges absolutely if this quantity is less than unity but diverges if it is greater than unity. It is particularly useful in connection with power series.
The test
The root test was developed first by Augustin-Louis Cauchy and so is sometimes known as the Cauchy root test or Cauchy's radical test. For a series
:\sum_{n=1}^\infty a_n.
the...
Reopening and closing question about Pascal's rule for binomial coefficients
Nov '1127
