# \sum_(n=1)^\inftyn/(n^4+1)?

## I would use Comparison Test but apparently we have to go with the Integral Test. Any tips (it's apparently only supposed to take 2-3 steps)? (btw, I assume this is an infinite series, but my terminology is rusty)

Apr 3, 2018

Converges by the Integral Test.

#### Explanation:

The Comparison Test would be the easiest option, but you can also use the Integral Test:

The Integral Test tells us if we have ${\sum}_{n = a}^{\infty} {a}_{n}$, we let $f \left(n\right) = {a}_{n} ,$ and create a new function $f \left(x\right) .$

If $f \left(x\right)$ is positive, continuous, and (ultimately) decreasing on $\left[a , \infty\right) ,$ we can take ${\int}_{a}^{\infty} f \left(x\right) \mathrm{dx}$. If that integral converges, so does the series ${\sum}_{n = a}^{\infty} {a}_{n} ,$ if it diverges, so does the series.

Let $f \left(x\right) = \frac{x}{{x}^{4} + 1}$

This is certainly continuous and positive on $\left[1 , \infty\right)$. It must also be decreasing (at least ultimately) to use the Integral Test. We'll use the First Derivative Test to prove the decreasing nature of this function, though it's fairly obvious at a glance.

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

$\frac{{x}^{4} + 1 - 4 {x}^{4}}{{x}^{4} + 1} ^ 2 = 0$

$- 3 {x}^{4} + 1 = 0$

$3 {x}^{4} = 1$

${x}^{4} = \frac{1}{3} , x = \pm \sqrt[4]{\frac{1}{3}}$ are critical values. The negative one is not in $\left[1 , \infty\right)$ so it's irrelevant to us. Let's see if $f \left(x\right)$ is decreasing on $\left(\frac{1}{\sqrt[4]{3}} , \infty\right)$:

$f ' \left(1\right) = \frac{- 3 + 1}{2} ^ 2 < 0$

So, the function is decreasing, we're good to use the integral test.

Let us first find the general indefinite integral:

$\int \frac{x}{{x}^{4} + 1} \mathrm{dx} = \int \frac{x}{{\left({x}^{2}\right)}^{2} + 1} \mathrm{dx}$

$u = {x}^{2} , \mathrm{du} = 2 x \mathrm{dx} , \frac{1}{2} \mathrm{du} = x \mathrm{dx}$

$\frac{1}{2} \int \frac{\mathrm{du}}{{u}^{2} + 1} = \frac{1}{2} \arctan \left(u\right) = \frac{1}{2} \arctan \left({x}^{2}\right)$

Now, we want

${\int}_{1}^{\infty} \frac{x}{{x}^{4} + 1} \mathrm{dx}$

So:

${\lim}_{t \to \infty} {\int}_{1}^{t} \frac{x}{{x}^{4} + 1} \mathrm{dx} = {\lim}_{t \to \infty} \left(\frac{1}{2} \arctan {x}^{2} {|}_{1}^{t}\right)$

$= {\lim}_{t \to \infty} \left(\frac{1}{2} \arctan t - \frac{1}{2} \arctan 1\right) = \left(\frac{1}{2} \left(\frac{\pi}{2}\right) - \frac{1}{2} \left(\frac{\pi}{4}\right)\right) = \frac{\pi}{4} - \frac{\pi}{8} = \frac{\pi}{8}$

Since the improper integral has a finite value, it converges.

Then, by the Integral Test, so does the series.