Let's define #f(n)=1/root(5)n#
We can verify that #f(n)# is ultimately decreasing (in fact, it is always decreasing). As we increase #n,# the numerator increases, causing the overall fraction to decrease.
Moreover,
#lim_(n->∞)f(n)=lim_(n->∞)1/root(5)n=1/∞=0#.
So, the two prerequisites for using the integral test are met.
Then,
#f(x)=1/root(5)x#.
This is important. We cannot stay in terms of #n# when wishing to integrate.
Let's now calculate
#int_1^∞1/root(5)xdx#
#int_1^∞1/root(5)xdx=int_1^∞1/x^(1/5)dx#
#int_1^∞1/x^(1/5)dx=lim_(t->∞)int_1^t1/x^(1/5)dx=lim_(t->∞)(5/4x^(4/5))# Where #5/4x^(4/5)# is evaluated from #1# to #t#:
#lim_(t->∞)5/4t^(4/5)-5/4=∞#.
So, since the integral of #f(x)# is divergent, it follows that the series containing #f(n)# as the sequence of summands is divergent.
