Question

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

Answer 1

`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 = (3k^2-3)/(k^5+1)`

Let's think about the end behaviour of `a_n`.

For large `n`, the numerator `3k^2-3` acts like `3k^2`.

Also, for large `n`, the denominator `k^5+1` acts like `k^5`.

So, for large `n`, `a_n` acts like `(3k^2)/k^5 = 3/k^3`.

Let `b_n= 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= 1/x^3` is convergent, so

`a_n = (3k^2-3)/(k^5+1)` is also convergent.