`sum_(n=1)^oon/(n^4+1)`

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^4+1)`

Consider `f(x)=x/(x^4+1)`

Refer to the attached...

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`sum_(n=1)^oon/(n^4+1)`

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^4+1)`

Consider `f(x)=x/(x^4+1)`

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

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

We can apply integral test , since the function satisfies the conditions for the integral test.

Now let's determine whether the corresponding improper integral `int_1^oox/(x^4+1)dx` converges or diverges.

`int_1^oox/(x^4+1)dx=lim_(b->oo)int_1^bx/(x^4+1)dx`

Let's first evaluate the indefinite integral `intx/(x^4+1)dx`

Apply integral substitution:`u=x^2`

`=>du=2xdx`

`intx/(x^4+1)dx=int1/(u^2+1)(du)/2`

Take the constant out and use common integral:`int1/(x^2+1)dx=arctan(x)+C`

`=1/2arctan(u)+C`

Substitute back `u=x^2`

`=1/2arctan(x^2)+C`

`int_1^oox/(x^4+1)dx=lim_(b->oo)[1/2arctan(x^2)]_1^b`

`=lim_(b->oo)[1/2arctan(b^2)]-[1/2arctan(1^2)]`

`=lim_(b->oo)[1/2arctan(b^2)]-[1/2arctan(1)]`

`=lim_(b->oo)[1/2arctan(b^2)]-[1/2(pi/4)]`

`=lim_(b->oo)[1/2arctan(b^2)]-pi/8`

`=1/2lim_(b->oo)arctan(b^2)-pi/8`

Now `lim_(b->oo)(b^2)=oo`

`=1/2(pi/2)-pi/8` [by applying the common limit:`lim_(u->oo)arctan(u)=pi/2` ]

`=pi/4-pi/8`

`=pi/8`

Since the integral `int_1^oox/(x^4+1)dx` converges, we conclude from the integral test that the series converges.