How do you use the integral test to determine whether the following series converge of diverge #sum n/((n^2+1)^2)# from n=1 to infinity? Thanks for the help !!! I have no idea on how to do these questions?

2 Answers
Jun 12, 2015

Answer:

The series converges.

Explanation:

Let #f(x) = x/(x^2+1)^2#.

In order to use the intergral test, #f(x)# must be positive and decreasing on #[p;+oo[#.

Let's study the sign of #f(x)# :
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#f(x)# is positive on #]0;+oo[#.

Let's study the slope of #f(x)# :

To find the derivative of #f(x)#, we will use the formula :

#((h(x))/(g(x)))' = (h'(x)g(x) - h(x)g'(x))/g^2(x)#

#f'(x) = ((x^2+1)^2-4x^2(x^2+1))/(x^2+1)^4 = (-3x^2+1)/(x^2+1)^3#
enter image source here
#f(x)# is decreasing on #]1/sqrt(3);+oo[#

We want to use the integral test for n=1 to infinity. Since #f(x)# is positive and decreasing on #]1/sqrt(3);+oo[#, it is also true for #[1;+oo[#.

#f(x) = x/(x^2+1)^2 = x(x^2+1)^(-2)#

To find the integral of #f(x)#, we will use the formula :

#inth'(x)h^n(x)dx = 1/(n+1)h^(n+1)(x)#

#intf(x)dx = intx(x^2+1)^(-2)dx = 1/2int2x(x^2+1)^(-2)dx#

#= 1/2inth'(x)*h^(-2)(x)dx#, where #h(x) = (x^2+1)#

#=1/2 * (1/((-2)+1)h^(-2+1)(x)) = -1/2h^(-1)(x) #

# = -1/2(x^2+1)^-1 = -1/(2(x^2+1)) = F(x)#

The series converges if #int_(1)^(+oo)f(x)dx# exists.

#int_(1)^(+oo)f(x)dx = [F(x)]_(1)^(+oo) = ''F(+oo)'' - F(1)#

#= 0 - F(1) = -(-1/(1+1)^2) = +1/4#.

The series converges.

Jun 12, 2015

The integral test just says, basically:

By taking the integral of a positive, decreasing function #a_n# of some series #sum a_n# that works within the boundaries #[k, oo]#, if the integral is finite, the sum converges.

#n/(n^2 + 1)^2# is obviously decreasing since it's basically #n/n^4 = 1/n^3#, which decreases as #n->oo#. It's also certainly positive if #n > 0#. Both conditions are satisfied.

So, integrate #n/(n^2 + 1)^2# from #1# to #oo#. Just replace #n# with #x#.

#sum_(n=1)^(oo) n/(n^2 + 1)^2# vs. #int_(1)^(oo) x/(x^2 + 1)^2dx#

Let:
#u = x^2 + 1#
#du = 2xdx#

#=> 1/2int_(1)^(oo) (2x)/(x^2 + 1)^2dx#

#= 1/2int_(1)^(oo) 1/(u^2)du#

#= 1/2 [-1/u]|_a^b#

#=> 1/2 (-1/(x^2+1))|_1^oo#

#= 1/2 [(-1/(oo^2+1)) - (-1/(1^2+1))]#

#= 1/2 [(0) - (-1/(2))] = 1/4#

The integral is finite, and therefore the series converges (#oo^2 = oo, 1/(oo) = 0#).

This is really just using the idea that an integral over an interval is just the accumulation of an infinite number of thin intervals #dn#.