How do you use the integral test to determine the convergence or divergence of #Sigma 1/root5(n)# from #[1,oo)#?

1 Answer
Mar 21, 2018

Diverges.

Explanation:

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.