Show all work in determining if the following series or integrals are absolutely convergent, conditionally convergent, or divergent. If divergent, tell how (approaches infinity, -infinity, or oscillates).
2) `int_0^oo tan^-1xdx`
3) `int_0^oo tan^3xdx`
4) `int_0^oo x^2/(x^2-5x+6)dx`
1) To determine this, let's use the ratio test:
`lim_(n->oo)|a_(n+1)/a_n| < 1`
Here, we can ignore absolute value because the result of the expression will be positive for all n. So, putting in our formula:
`lim_(n->oo)((prod_1^(n+1) (2n-1))/(5^(n+1)(n+1)!))/((prod_1^n(2n-1))/(5^n n!))`
Now, we'll take some terms out to simplify:
`lim_(n->oo) ((2n+1)/(5(n+1)) (prod_1^n(2n-1))/(5^n n!))/((prod_1^n(2n-1))/(5^n n!))`
Now, the giant term cancels itself out, thankfully!
`lim_(n->oo) (2n+1)/(5n+5) = 2/5 < 1`
Therefore, by the ratio test, this series converges.
2) To find whether this integral converges, we can see whether the limit of the function approaches 0 as x approaches infinity. This is fairly easy because considering the inverse tangent will be the tangent reflected over the line y=x, the limit as x approaches infinity will be the vertical asymptote of tan(x): `pi/2`. So, we get:
`lim_(x->oo) tan^-1x = pi/2`
So, as we sum to infinity, we get no oscillation, and the integral diverges to `+oo`.
3) This integral should diverge in an oscillating fashion, knowing what the graph of `tan^3x` should look like. However, let's see what we can do to prove it. Start with the integral:
`X= int_0^oo tan^3x dx`
We can integrate this by using trigonometric identities:
`int_0^oo tan^2x tanx dx = int_0^oo (sec^2x - 1)tanx dx`
`= int_0^oo sec^2xtanx dx- int_0^oo tanx dx `
The first integral can be solved by noting that `sec^2x tanx` is the chain-rule-based derivative of `(tan^2x)/2`. The second integral can be found by using an integral table:
`= ((tan^2x)/2 - log|secx|)|_0^oo`
Now, we can try to take the limit as x approaches infinity:
`lim_(x->oo) (tan^2x)/2 - log|secx|`
Here is the problem, though. Neither term converges!
Clearly, this integral diverges and oscillates between 0 and `+oo`.
4)This integral can be shown to diverge in the same fashion as (2). Let's examine the limit of the function as it approaches infinity:
There is a way to do this by getting rid of the terms that are not of the greatest degree in the numerator and denominator. However, let's make a different approach by dividing numerator and denominator by `x^2`:
`lim_(x->oo) 1/(1-5/x + 6/x^2)`
As x approaches infinity, the `5/x` and `6/x^2` terms approach 0 (For this reason, we only need to consider terms of the greatest degree in the original limit, in case you were wondering). We end up with the following:
`lim_(x->oo) 1/1 = 1`
Because the limit is not 0, and is stable at 1, this integral diverges to `+oo`.