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.