# 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...

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).

1)`sum_(n=0)^oo(1xx3xx5xx...xx(2n-1))/(5^n n!)`

2) `int_0^oo tan^-1xdx`

3) `int_0^oo tan^3xdx`

4) `int_0^oo x^2/(x^2-5x+6)dx`

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txmedteach | High School Teacher | (Level 3) Associate Educator

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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:

`lim_(x->oo) x^2/(x^2-5x+6)`

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`.