Our best bet is using the Divergence Test to determine the divergence of the series.
This entails taking #lim_(x->oo)((x+4)/x)^x#. If this limit is non-zero, the series must diverge (the converse is not true, if the limit is zero, it is inconclusive).
Plugging in #oo# right away yields #1^oo,# an indeterminate form, so some manipulation is needed.
Let #y=((x+4)/x)^x#
Then,
#lny=ln((x+4)/x)^x#
#lny=xln((x+4)/x)=(ln((x+4)/x))/(1/x)#
Now, we can take
#lim_(x->oo)lny=lim_(x->oo)(ln((x+4)/x))/(1/x)=ln(1)/0=0/0#, indeterminate, so we will be best off using l'Hospital's rule.
(The sequence you have provided is already in terms of #x,# not in terms of #n,# so I'm assuming we do not need to redefine a new differentiable function which l'Hospital's Rule can be used on)
#lim_(x->oo)(ln((x+4)/x))/(1/x)=lim_(x->oo)(ln(x+4)-ln(x))/x^-1#, splitting up the log allows for easier differentiation.
#=lim_(x->oo)(1/(x+4)-1/x)/(-x^-2)=lim_(x->oo)(x-x-4)/((x)(x+4)(-x^-2))=lim_(x->oo)(-4)/((x^2+4x)(-x^-2))=lim_(x->oo)(-4)/(-1-4/x)=4#
Thus, #lim_(x->oo)lny=4#, meaning that #lim_(x->oo)y=lim_(x->oo)e^(lny)=e^4 ne 0# and so the series diverges.
