# Is the series \sum_(n=1)^\infty\tan^-1(1/n) absolutely convergent, conditionally convergent or divergent?

## (Use the appropriate test.)

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#### Explanation

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VNVDVI Share
Apr 23, 2018

Diverges by Limit Comparison Test.

#### Explanation:

The Integral Test can be used; however, this is a very ugly function to integrate.

So, let's use the Limit Comparison Test instead, which tells us if we have some sequence ${b}_{n} \ge 0$ for all $n$ and we know $\sum {b}_{n}$ converges or diverges, then we let

$c = {\lim}_{n \to \infty} {a}_{n} / {b}_{n}$

And if $c > 0 \ne \infty$, then both series either converge or diverge.

Let's compare the given series to ${\sum}_{n = 1}^{\infty} \frac{1}{n}$. We know this harmonic series, in the form $\sum \frac{1}{n} ^ p$ with $p = 1$, diverges by the $p -$series test.

So, let's use the Limit Comparison Test:

$c = {\lim}_{n \to \infty} \arctan \frac{\frac{1}{n}}{\frac{1}{n}} = \arctan \frac{0}{0} = \frac{0}{0}$

This is an indeterminate form and we need to use l'Hospital's Rule.

Temporarily rewrite in terms of $x$, a differentiable variable, as $n$ is not really differentiable, and differentiate numerator and denominator:

First, let's consider $\frac{d}{\mathrm{dx}} \arctan \left(\frac{1}{x}\right) :$

In general, the derivative of the arctangent function is

$\frac{d}{\mathrm{dx}} \arctan x = \frac{1}{1 + {x}^{2}}$

So, for our function, using the Chain Rule, we'll temporarily say that $u = \frac{1}{x}$, so, still differentiating with respect to $x$ we get

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

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \frac{1}{x} = \frac{d}{\mathrm{dx}} {x}^{-} 1 = - {x}^{-} 2 = - \frac{1}{x} ^ 2$

${u}^{2} = {\left(\frac{1}{x}\right)}^{2}$

Thus,

$\frac{d}{\mathrm{dx}} \arctan \left(\frac{1}{x}\right) = - \frac{1}{x} ^ 2 \cdot \frac{1}{1 + {\left(\frac{1}{x}\right)}^{2}}$

${\lim}_{x \to \infty} \arctan \frac{\frac{1}{x}}{\frac{1}{x}} = {\lim}_{x \to \infty} \frac{\cancel{- \frac{1}{x} ^ 2 \cdot} \frac{1}{1 + {\left(\frac{1}{x}\right)}^{2}}}{\cancel{- \frac{1}{x} ^ 2}}$

$= {\lim}_{x \to \infty} \frac{1}{1 + {\left(\frac{1}{x}\right)}^{2}} = \frac{1}{1} = 1 > 0 \ne \infty$

Thus, ${\sum}_{n = 1}^{\infty} \arctan \left(\frac{1}{n}\right)$ must also diverge by the Limit Comparison Test.

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#### Explanation

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Apr 20, 2018

Divergent

#### Explanation:

$\setminus {\tan}^{-} 1 \left(\frac{1}{n}\right)$ is that angle of a right-angled triangle with a unit opposite and an adjacent equal to $n$. As $n$ increases, that angle decreases.

$\implies$ Use the integral test.

Setting it up for IBP (with the parenthesized hint):

$I = {\int}_{1}^{\infty} \setminus \setminus {\tan}^{-} 1 \left(\frac{1}{x}\right) \setminus d \left(x\right)$

$= {\left(x {\tan}^{- 1} \left(\frac{1}{x}\right)\right)}_{1}^{\infty} - {\int}_{1}^{\infty} \setminus x \setminus d \left({\tan}^{-} 1 \left(\frac{1}{x}\right)\right)$

Well-known:

• $d \left({\tan}^{- 1} u\right) = \frac{1}{1 + {u}^{2}} \setminus \mathrm{du}$

• $\mathrm{du} = d \left(\frac{1}{x}\right) = - \frac{1}{x} ^ 2 \setminus \mathrm{dx}$

$\implies I = {\left(x {\tan}^{- 1} \left(\frac{1}{x}\right)\right)}_{1}^{\infty} - {\int}_{1}^{\infty} \setminus x \setminus \frac{1}{1 + {\left(\frac{1}{x}\right)}^{2}} \setminus \left(- \frac{1}{x} ^ 2\right) \setminus \mathrm{dx}$

$= {\left(x {\tan}^{- 1} \left(\frac{1}{x}\right)\right)}_{1}^{\infty} + {\int}_{1}^{\infty} \setminus \frac{x}{{x}^{2} + 1} \setminus \mathrm{dx}$

$= {\left(x {\tan}^{- 1} \left(\frac{1}{x}\right)\right)}_{1}^{\infty} + \frac{1}{2} {\int}_{1}^{\infty} \setminus d \left(\ln | {x}^{2} + 1 |\right)$

$= {\left(x {\tan}^{- 1} \left(\frac{1}{x}\right) + \setminus \frac{1}{2} \ln | {x}^{2} + 1 | \setminus\right)}_{1}^{x \to \infty}$

The second term clearly tends toward $\infty$, ie it diverges .

For the first term, we can apply L'Hôpital because:

${\lim}_{x \to \infty} x {\tan}^{- 1} \left(\frac{1}{x}\right) = {\lim}_{x \to \infty} \frac{{\tan}^{- 1} \left(\frac{1}{x}\right)}{\frac{1}{x}} \left(= \frac{0}{0}\right)$

Applying L'H:

$= {\lim}_{x \to \infty} \frac{\frac{1}{1 + {\left(\frac{1}{x}\right)}^{2}} \cdot \left(- \frac{1}{x} ^ 2\right)}{- \frac{1}{x} ^ 2} = 1$

The integral diverges: and if the integral diverges, then the series does so as well.

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