In order to find out which test to use, the best way to do it is try more of them or simply try whichever you think will work.
In this example, it seems as if all terms are bigger than #1#. In order to prove this, we'll use the limit test.
Let #L = lim_(n->oo) ((n+2)/n)^n#.
#L = lim_(n->oo) (1+2/n)^n#
Take the natural logarithm of both sides.
#lnL = lim_(n->oo) nln(1+2/n)#
Since #ln(1+2/n)# approaches zero and #n# approaches infinity, we have a #oo * 0# situation. We can rewrite it as :
#lnL = lim_(n->oo) ln(1+2/n)/(1/n)#
This is a #0/0# case and we can use #color(red)("L'Hôpital's rule")#.
#lnL = lim_(n->oo) (d/(dn)ln(1+2/n))/(d/(dn)1/n#
The derivative of #lnf(x)#, for any function #f#, is #(f')/f#.
#d/(dn) ln(1+2/n) = (d/(dn)(1+2/n))/(1+2/n)#
#d/(dn) ln(1+2/n) = (2d/(dn)1/n)/(1+2/n)#
#1/n = n^(-1) => color(red)(d/(dn)1/n = -1/n^2#
#color(red)(d/(dn) ln(1+2/n) = -2/(n(n+2))#
By substituting these into the limit, we have
#ln L = lim_(n->oo)-2/(n(n+2)) * (-n^2) = lim_(n->oo)(2n^2)/(n^2+2n)#
#ln L = lim_(n->oo)(color(red)(2n))/(color(red)n+2)#
Because we are talking about numbers approaching infinity, #n/(n+2) -> 1#, as the #2# is having a smaller and smaller effect over it.
#ln L = lim_(n->oo) 2 * color(red)(n/(n+2)) = 2 * 1 = 2#
#e^(ln L) = e^2#
#L = e^2#
The limit test states that an infinite sum #sum_n^oo a_n# coverges only if #lim_(k->oo) a_k = 0#.
In our case, #lim_(k->oo) a_k = e^2 != 0#, therefore the series diverges.
