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

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Answer 1 · 2016-11-23T21:32:41

$sum_ (n=1)^oo(-1)^(n+1)sechn$ is convergent.

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

$sech(n)=2/(e^n+e^(-n))$ but

$1/(e^(n+1)+e^(-n-1)) < 1/(e^n+e^(-n))$

because

$1+e^(-2n) < e(1+e^(-2(n+1)))$

note that for $n > 0$

$1+e^(-2n) < 2$ and $e = 2.71$

so $sech(n+1) < sech(n)$ so

$sum_ (n=1)^oo(-1)^(n+1)sechn$ is convergent.

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