Determine whether the series sum_(n=1)^oo (2n^2 +3n)/sqrt(5+n^5)n=12n2+3n5+n5 is convergent or divergent. How do i tell which comparison test to use?

1 Answer
Mar 22, 2017

The series:

sum_(n=1)^oo (2n^2+3n)/sqrt(5+n^5)n=12n2+3n5+n5

is divergent.

Explanation:

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)n=1an=n=12n2+3n5+n5

We can see that the numerator is of order n^2n2 and the denominator is of order n^(5/2)n52.

For n->oon then the sequence tends to zero with order n^(-1/2)n12 and thus the series will not converge because:

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

is convergent only for p>1p>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) n=1bn=n=11n=n=11n12

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.