# How do you use the comparison test for sum (3k^2-3) / ((k^5)+1) for n=1 to n=oo?

Aug 11, 2015

color(red)(sum_(n=1)^∞ (3k^2-3)/(k^5+1)" is convergent").

#### Explanation:

sum_(n=1)^∞ (3k^2-3)/(k^5+1)

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 ${a}_{n} = \frac{3 {k}^{2} - 3}{{k}^{5} + 1}$

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

For large $n$, the numerator $3 {k}^{2} - 3$ acts like $3 {k}^{2}$.

Also, for large $n$, the denominator ${k}^{5} + 1$ acts like ${k}^{5}$.

So, for large $n$, ${a}_{n}$ acts like $\frac{3 {k}^{2}}{k} ^ 5 = \frac{3}{k} ^ 3$.

Let ${b}_{n} = \frac{1}{k} ^ 3$

Then lim_(n→∞)(a_n/b_n) = lim_(n→∞)((3k^2-3)/(k^5+1))/(1/k^3)= lim_(n→∞)((3k^2-3)k^3)/(k^5+1) = lim_(n→∞)( (3k^5 +k^3)/(k^5+1)) = lim_(n→∞)( (3+1/k^2)/(1+1/k^5)) =3

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

But ${b}_{n} = \frac{1}{x} ^ 3$ is convergent, so

${a}_{n} = \frac{3 {k}^{2} - 3}{{k}^{5} + 1}$ is also convergent.