The terms

n$3$

n$3+8$n$2+8$

approach $1$ as n $->\backslash$infty $.$ But the terms

$(1)$n $+5$ n$3$

n$3+8$n$2+8$

alternate in sign. Therefore, the terms

a$1,$ a$3,$ a$5,...$

approach

$,$ while the terms

a$2,$ a$4,$ a$6,...$

approach

$.$