Question

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

Answer 1

`sum_(n=1)^oo 5/(6n^2+n-1)` is convergent.

Explanation:

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.