`sum_(n=1)^oo n/(n^4+2n^2+1)` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.

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gsarora17 eNotes educator| Certified Educator

`sum_(n=1)^oon/(n^4+2n^2+1)`

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

For the given series `a_n=n/(n^4+2n^2+1)`

Consider `f(x)=x/(x^4+2x^2+1)`

`f(x)=x/(x^2+1)^2`

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

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

Apply the quotient rule to find the derivative,

`f'(x)=((x^2+1)^2d/dx(x)-xd/dx(x^2+1)^2)/(x^2+1)^4`

`f'(x)=((x^2+1)^2-x(2(x^2+1)2x))/(x^2+1)^4`

`f'(x)=((x^2+1)(x^2+1-4x^2))/(x^2+1)^4`

`f'(x)=(-3x^2+1)/(x^2+1)^3`

`f'(x)=-(3x^2-1)/(x^2+1)^3<0`

Since the function satisfies the conditions for the integral test, we can apply the integral test.

Now let's determine the convergence or divergence of the improper integral as follows:

`int_1^oox/(x^2+1)^2dx=lim_(b->oo)int_1^bx/(x^2+1)^2dx`

Let's first evaluate the indefinite integral `intx/(x^2+1)^2dx`

Apply integral substitution:`u=x^2+1`

`=>du=2xdx`

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

`=1/2int1/u^2du`

Apply the power rule,

`=1/2(u^(-2+1)/(-2+1))`

`=-1/(2u)`

Substitute back `u=(x^2+1)`

`=-1/(2(x^2+1))+C`  where C is a constant

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

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

`=-1/2[0-1/2]`

`=1/4`

Since the integral `int_1^oox/(x^4+2x^2+1)dx` converges, we conclude from the integral test that the series `sum_(n=1)^oon/(n^4+2n^2+1)` converges.

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