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

(Use the appropriate test.)

(Use the appropriate test.)

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Apr 23, 2018

Answer:

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.

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Apr 20, 2018

Answer:

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.

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