Determine whether the series # sum_(n=1)^oo (2n^2 +3n)/sqrt(5+n^5)# is convergent or divergent. How do i tell which comparison test to use?

We need to determine the convergence of the series:

#sum_(n=1)^oo a_n = sum_(n=1)^oo (2n^2+3n)/sqrt(5+n^5)#

We can see that the numerator is of order #n^2# and the denominator is of order #n^(5/2)#.

For #n->oo# then the sequence tends to zero with order #n^(-1/2)# and thus the series will not converge because:

#sum_(n=1)^oo n^(-p)#

is convergent only for #p>1# (p-series test).

To demonstrate this formally, consider the series:

#sum_(n=1)^oo b_n = sum_(n=1)^oo 1/sqrt(n) = sum_(n=1)^oo 1/n^(1/2) #

as stated above this series is divergent based on the p-series test.

Now evaluate the limit:

#lim_(n->oo) a_n/b_n = lim_(n->oo) ((2n^2+3n)/sqrt(5+n^5))/(1/sqrt(n))#

#lim_(n->oo) a_n/b_n = lim_(n->oo) (n sqrt(n)(2n+3))/sqrt(5+n^5)#

#lim_(n->oo) a_n/b_n = lim_(n->oo) sqrt( (n^3(2n+3)^2)/(5+n^5)#

#lim_(n->oo) a_n/b_n = lim_(n->oo) sqrt( (n^3(4n^2+12n+9))/(5+n^5)#

#lim_(n->oo) a_n/b_n = lim_(n->oo) sqrt( (4n^5+12n^3+9n^3)/(5+n^5) )=2#

As the limit exists and is finite, based on the limit comparison test also the first series is divergent.