Question

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

Answer 1

`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 = (2x^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 `(2x^4)/x^5 = 2/x`.

Let `b_n= 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= 1/x` is divergent, so

`a_n = x^4/(x^5-10)` is also divergent.