# How do you determine the convergence or divergence of Sigma (-1)^(n+1)sechn from [1,oo)?

Nov 23, 2016

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} \sech n$ is convergent.

#### Explanation:

For alternating series, if the series is absolutely convergent, then it is also convergent. Now

$\sech \left(n\right) = \frac{2}{{e}^{n} + {e}^{- n}}$ but

$\frac{1}{{e}^{n + 1} + {e}^{- n - 1}} < \frac{1}{{e}^{n} + {e}^{- n}}$

because

$1 + {e}^{- 2 n} < e \left(1 + {e}^{- 2 \left(n + 1\right)}\right)$

note that for $n > 0$

$1 + {e}^{- 2 n} < 2$ and $e = 2.71$

so $\sech \left(n + 1\right) < \sech \left(n\right)$ so

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} \sech n$ is convergent.