How do you use the integral test to determine whether the following series converge of diverge sum n/((n^2+1)^2) from n=1 to infinity? Thanks for the help !!! I have no idea on how to do these questions?

Jun 12, 2015

The series converges.

Explanation:

Let $f \left(x\right) = \frac{x}{{x}^{2} + 1} ^ 2$.

In order to use the intergral test, $f \left(x\right)$ must be positive and decreasing on [p;+oo[.

Let's study the sign of $f \left(x\right)$ :

$f \left(x\right)$ is positive on ]0;+oo[.

Let's study the slope of $f \left(x\right)$ :

To find the derivative of $f \left(x\right)$, we will use the formula :

$\left(\frac{h \left(x\right)}{g \left(x\right)}\right) ' = \frac{h ' \left(x\right) g \left(x\right) - h \left(x\right) g ' \left(x\right)}{g} ^ 2 \left(x\right)$

$f ' \left(x\right) = \frac{{\left({x}^{2} + 1\right)}^{2} - 4 {x}^{2} \left({x}^{2} + 1\right)}{{x}^{2} + 1} ^ 4 = \frac{- 3 {x}^{2} + 1}{{x}^{2} + 1} ^ 3$

$f \left(x\right)$ is decreasing on ]1/sqrt(3);+oo[

We want to use the integral test for n=1 to infinity. Since $f \left(x\right)$ is positive and decreasing on ]1/sqrt(3);+oo[, it is also true for [1;+oo[.

$f \left(x\right) = \frac{x}{{x}^{2} + 1} ^ 2 = x {\left({x}^{2} + 1\right)}^{- 2}$

To find the integral of $f \left(x\right)$, we will use the formula :

$\int h ' \left(x\right) {h}^{n} \left(x\right) \mathrm{dx} = \frac{1}{n + 1} {h}^{n + 1} \left(x\right)$

$\int f \left(x\right) \mathrm{dx} = \int x {\left({x}^{2} + 1\right)}^{- 2} \mathrm{dx} = \frac{1}{2} \int 2 x {\left({x}^{2} + 1\right)}^{- 2} \mathrm{dx}$

$= \frac{1}{2} \int h ' \left(x\right) \cdot {h}^{- 2} \left(x\right) \mathrm{dx}$, where $h \left(x\right) = \left({x}^{2} + 1\right)$

$= \frac{1}{2} \cdot \left(\frac{1}{\left(- 2\right) + 1} {h}^{- 2 + 1} \left(x\right)\right) = - \frac{1}{2} {h}^{- 1} \left(x\right)$

$= - \frac{1}{2} {\left({x}^{2} + 1\right)}^{-} 1 = - \frac{1}{2 \left({x}^{2} + 1\right)} = F \left(x\right)$

The series converges if ${\int}_{1}^{+ \infty} f \left(x\right) \mathrm{dx}$ exists.

${\int}_{1}^{+ \infty} f \left(x\right) \mathrm{dx} = {\left[F \left(x\right)\right]}_{1}^{+ \infty} = ' ' F \left(+ \infty\right) ' ' - F \left(1\right)$

$= 0 - F \left(1\right) = - \left(- \frac{1}{1 + 1} ^ 2\right) = + \frac{1}{4}$.

The series converges.

Jun 12, 2015

The integral test just says, basically:

By taking the integral of a positive, decreasing function ${a}_{n}$ of some series $\sum {a}_{n}$ that works within the boundaries $\left[k , \infty\right]$, if the integral is finite, the sum converges.

$\frac{n}{{n}^{2} + 1} ^ 2$ is obviously decreasing since it's basically $\frac{n}{n} ^ 4 = \frac{1}{n} ^ 3$, which decreases as $n \to \infty$. It's also certainly positive if $n > 0$. Both conditions are satisfied.

So, integrate $\frac{n}{{n}^{2} + 1} ^ 2$ from $1$ to $\infty$. Just replace $n$ with $x$.

${\sum}_{n = 1}^{\infty} \frac{n}{{n}^{2} + 1} ^ 2$ vs. ${\int}_{1}^{\infty} \frac{x}{{x}^{2} + 1} ^ 2 \mathrm{dx}$

Let:
$u = {x}^{2} + 1$
$\mathrm{du} = 2 x \mathrm{dx}$

$\implies \frac{1}{2} {\int}_{1}^{\infty} \frac{2 x}{{x}^{2} + 1} ^ 2 \mathrm{dx}$

$= \frac{1}{2} {\int}_{1}^{\infty} \frac{1}{{u}^{2}} \mathrm{du}$

$= \frac{1}{2} \left[- \frac{1}{u}\right] {|}_{a}^{b}$

$\implies \frac{1}{2} \left(- \frac{1}{{x}^{2} + 1}\right) {|}_{1}^{\infty}$

$= \frac{1}{2} \left[\left(- \frac{1}{{\infty}^{2} + 1}\right) - \left(- \frac{1}{{1}^{2} + 1}\right)\right]$

$= \frac{1}{2} \left[\left(0\right) - \left(- \frac{1}{2}\right)\right] = \frac{1}{4}$

The integral is finite, and therefore the series converges (${\infty}^{2} = \infty , \frac{1}{\infty} = 0$).

This is really just using the idea that an integral over an interval is just the accumulation of an infinite number of thin intervals $\mathrm{dn}$.