# How do you use the limit comparison test for sum (2x^4)/(x^5+10) n=1 to n=oo?

Aug 11, 2015

color(red)(sum_(n=1)^∞ (2x^4)/(x^5-10)" is divergent").

sum_(n=1)^∞ (2x^4)/(x^5+10)

The limit comparison test states that if ${a}_{n}$ and ${b}_{n}$ are series with positive terms and if lim_(n→∞) (a_n)/(b_n) is positive and finite, then either both series converge or both diverge.

Let ${a}_{n} = \frac{2 {x}^{4}}{{x}^{5} + 10}$

Let's think about the end behaviour of ${a}_{n}$.

For large $n$, the denominator ${x}^{5} + 10$ acts like ${x}^{5}$.

So, for large $n$, ${a}_{n}$ acts like $\frac{2 {x}^{4}}{x} ^ 5 = \frac{2}{x}$.

Let ${b}_{n} = \frac{1}{x}$

Then lim_(n→∞)(a_n/b_n) = lim_(n→∞)( ((2x^4)/(x^5+10))/(1/x)) = lim_(n→∞)( (2x^4×x)/(x^5+10)) = lim_(n→∞)( (2x^5)/(x^5+10)) = lim_(n→∞)( 2/(1-1/x^5)) =2

The limit is both positive and finite, so either ${a}_{n}$ and ${b}_{n}$ are both divergent or both are convergent.

But ${b}_{n} = \frac{1}{x}$ is divergent, so

${a}_{n} = {x}^{4} / \left({x}^{5} - 10\right)$ is also divergent.