`1/4+2/7+3/12+.....+n/(n^2+3)+........`

We can write the series as `sum_(n=1)^oon/(n^2+3)`

The integral test is applicable if f is positive , continuous and decreasing function on infinite interval `[k,oo)` where `k>=1` and `a_n=f(x)` . Then the series `sum_(n=1)^ooa_n` converges or diverges if and only if the improper integral `int_1^oof(x)dx` converges or diverges.

For the given series `a_n=n/(n^2+3)`

Consider `f(x)=x/(x^2+3)`

Refer to the attached graph of the function. From the graph we observe that the function is positive, continuous and decreasing on the interval `[1,oo)`

We can also determine whether function is decreasing by finding the derivative f'(x) such that `f'(x)<0`  for `x>=1`

Now let's determine whether the corresponding improper integral `int_1^oox/(x^2+3)dx` converges or diverges.

`int_1^oox/(x^2+3)dx=lim_(b->oo)int_1^bx/(x^2+3)dx`

Let's first evaluate the indefinite integral `intx/(x^2+3)dx`

Apply integral substitution:`u=x^2+3`

`=>du=2xdx`

`intx/(x^2+3)dx=int1/u(du)/2`

Take the constant out and use the common integral:`int1/xdx=ln|x|`

`=1/2ln|u|+C`  where C is a constant

Substitute back `u=x^2+3`

`=1/2ln|x^2+3|+C`

`int_1^oox/(x^2+3)dx=lim_(b->oo)[1/2ln|x^2+3|]_1^b`

`=lim_(b->oo)1/2ln|b^2+3|-1/2ln|1^2+3|`

`=oo-1/2ln4`

`=oo`

Since the integral `int_1^oox/(x^2+3)dx` diverges, we can conclude from the integral test that the series diverges.

Images:

Image (1 of 1)