How do you test for convergence of #Sigma 5/(6n^2+n-1)# from #n=[1,oo)#?

As the terms of the series are positive we can use the integral test, using:

#f(x) = 5/(6x^2+x-1)#

We have that:

#f(x) > 0# for #x in [1,+oo)]#

#f'(x) = - frac (5(12x+1)) ((6x^2+x-1)^2) < 0 # for #x in [1,+oo]#, so the function is monotone decreasing in that interval.

#lim_(x->oo) 5/(6x^2+x-1) = 0#

#f(n) = 5/(6n^2+n-1)#

so all the hypotheses of the integral test are satisfied and we can calculate:

#int_1^oo (5dx)/(6x^2+x-1)#

using partial fractions:

#6x^2 +x -1 = (2x+1)(3x-1)#

#A/(2x+1) + B/(3x-1) = 5/((2x+1)(3x-1))#

#A(3x-1)+B(2x+1) = 5#

#(3A+2B)x^2 -(A-B) = 5#

#{ color(white) [ color(black) ((3A+2B = 0) ,( A-B = -5) ) color(white) ] color(black)#

#{ color(white) [ color(black) ((A=-2 ),( B=3) ) color(white) ] color(black)#

So:

#int_1^oo (5dx)/(6x^2+x-1) = int_1^oo (-2dx)/(2x+1) +int_1^oo (3dx)/(3x-1) = [ln(3x-1) - ln (2x+1)]_1^oo#

Using the properties of logarithms:

#ln(3x-1) - ln(2x+1) = ln ((3x-1)/(2x+1))#

so that:

#[ln(3x-1) - ln (2x+1)]_1^oo = [ln ((3x-1)/(2x+1))]_1^oo#

And as:
# lim_(x->oo) ln ((3x-1)/(2x+1)) = ln(3/2)#

#[ln ((3x-1)/(2x+1))]_1^oo = ln(3/2) - ln (2/3) = 2ln(3/2)#

Finally we have:

#int_1^oo (5dx)/(6x^2+x-1) = 2ln(3/2)#

and as the integral is convergent, so is the series.