Question

Is the series `\sum_(n=1)^\infty\tan^-1(1/n)` absolutely convergent, conditionally convergent or divergent?

(Use the appropriate test.)

Answer 1

Divergent

Explanation:

`\tan^-1(1/n)` is that angle of a right-angled triangle with a unit opposite and an adjacent equal to `n`. As `n` increases, that angle decreases.

`implies` Use the integral test.

Setting it up for IBP (with the parenthesized hint):

`I = int_1^oo \ \tan^-1(1/x) \ d(x)`

`= (x tan^(-1)(1/x))_1^oo - int_1^oo \ x \ d(tan^-1(1/x) )`

Well-known:

• `d(tan^(-1) u) = 1/(1+ u^2) \du`

• `du = d(1/x) = - 1/x^2\ dx`

`implies I = (x tan^(-1)(1/x))_1^oo - int_1^oo \ x \ 1/(1+ (1/x)^2) \ (- 1/x^2) \ dx`

`= (x tan^(-1)(1/x))_1^oo + int_1^oo \ x/(x^2+ 1) \ dx`

`= (x tan^(-1)(1/x))_1^oo + 1/2 int_1^oo \ d( ln |x^2+ 1| )`

`= (x tan^(-1)(1/x) + \ 1/2 ln |x^2+ 1| \ )_1^(x to oo)`

The second term clearly tends toward `oo`, ie it diverges .

For the first term, we can apply L'Hôpital because:

`lim_(x to oo) x tan^(-1)(1/x) = lim_(x to oo) (tan^(-1)(1/x))/(1/x) (= 0/0 )`

Applying L'H:

`= lim_(x to oo) (1/(1+(1/x)^2)* (- 1/x^2))/(-1/x^2) = 1`

The integral diverges: and if the integral diverges, then the series does so as well.

Answer 2

Diverges by Limit Comparison Test.

Explanation:

The Integral Test can be used; however, this is a very ugly function to integrate.

So, let's use the Limit Comparison Test instead, which tells us if we have some sequence `b_n>=0` for all `n` and we know `sumb_n` converges or diverges, then we let

`c=lim_(n->oo)a_n/b_n`

And if `c>0neoo`, then both series either converge or diverge.

Let's compare the given series to `sum_(n=1)^oo1/n`. We know this harmonic series, in the form `sum1/n^p` with `p=1`, diverges by the `p-`series test.

So, let's use the Limit Comparison Test:

`c=lim_(n->oo)arctan(1/n)/(1/n)=arctan0/0=0/0`

This is an indeterminate form and we need to use l'Hospital's Rule.

Temporarily rewrite in terms of `x`, a differentiable variable, as `n` is not really differentiable, and differentiate numerator and denominator:

First, let's consider `d/dxarctan(1/x):`

In general, the derivative of the arctangent function is

`d/dxarctanx=1/(1+x^2)`

So, for our function, using the Chain Rule, we'll temporarily say that `u=1/x`, so, still differentiating with respect to `x` we get

`(du)/dxarctan(u)=1/(1+u^2)*(du)/dx`

`(du)/dx=d/dx1/x=d/dxx^-1=-x^-2=-1/x^2`

`u^2=(1/x)^2`

Thus,

`d/dxarctan(1/x)=-1/x^2*1/(1+(1/x)^2)`

`lim_(x->oo)arctan(1/x)/(1/x)=lim_(x->oo)(cancel(-1/x^2*)1/(1+(1/x)^2))/(cancel(-1/x^2))`

`=lim_(x->oo)1/(1+(1/x)^2)=1/1=1>0neoo`

Thus, `sum_(n=1)^ooarctan(1/n)` must also diverge by the Limit Comparison Test.