# Identify whether the infinite series converge absolutely conditionally or dont sum_(n=1)^oo (-1)^(n+1) (n (arctan(n+1)-arctan (n)) (Apply Mean Value theorem to conclude)?

##### 1 Answer
Jul 25, 2018

The series:

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} n \left(\arctan \left(n + 1\right) - \arctan \left(n\right)\right)$

is conditionally convergent but not absolutely convergent.

#### Explanation:

Let ${a}_{n} = n \left(\arctan \left(n + 1\right) - \arctan \left(n\right)\right)$.

The Mean Value Theorem states that if $f \left(x\right)$ is continuous in $\left[a , b\right]$ and differentiable in $\left(a , b\right)$, then there is a point $\xi \in \left(a , b\right)$ such that:

$f ' \left(c\right) = \frac{f \left(b\right) - f \left(a\right)}{b - a}$

Let $f \left(x\right) = \arctan x$ and $a = n$, $b = n + 1$, then:

$\frac{1}{1 + {\xi}^{2}} = \arctan \left(n + 1\right) - \arctan \left(n\right)$

with $n < \xi < n + 1$. As for $x > 0$ the function $f ' \left(x\right) = \frac{1}{1 + {x}^{2}}$ is strictly decreasing, we have:

$\frac{1}{1 + {\left(n + 1\right)}^{2}} < \arctan \left(n + 1\right) - \arctan \left(n\right) < \frac{1}{1 + {n}^{2}}$

and multiplying by $n$:

$\left(1\right) \text{ } \frac{n}{1 + {\left(n + 1\right)}^{2}} < {a}_{n} < \frac{n}{1 + {n}^{2}}$

Now the series:

$\sum \frac{n}{1 + {\left(n + 1\right)}^{2}}$

is not convergent as we can see using the limit comparison test with the harmonic series. In fact:

${\lim}_{n \to \infty} \frac{\frac{n}{1 + {\left(n + 1\right)}^{2}}}{\frac{1}{n}} = {\lim}_{n \to \infty} {n}^{2} / \left({n}^{2} + 2 n + 2\right) = 1$

By direct comparison we can then conclude that also:

$\sum {a}_{n}$

is not convergent hence the series in analysis is not absolutely convergent.

On the other hand, let:

${b}_{n} = \frac{n}{1 + {\left(n + 1\right)}^{2}}$

${c}_{n} = \frac{n}{1 + {n}^{2}}$

We have that:

${b}_{n} > 0$ and ${c}_{n} > 0$

${\lim}_{n \to \infty} {b}_{n} = {\lim}_{n \to \infty} {c}_{n} = 0$

and:

${b}_{n} = g \left(n\right)$ with $g \left(x\right) = \frac{x}{1 + {\left(1 + x\right)}^{2}}$

$g \left(x\right) = \frac{x}{2 + 2 x + {x}^{2}}$

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

$\frac{\mathrm{dg}}{\mathrm{dx}} = - \frac{3 {x}^{2} - 2}{2 + 2 x + {x}^{2}} ^ 2 < 0$ for $x > 1$

hence: ${b}_{n + 1} < {b}_{n}$ and similarly:

${c}_{n} = h \left(n\right)$ with $h \left(x\right) = \frac{x}{1 + {x}^{2}}$

$\frac{\mathrm{dh}}{\mathrm{dx}} = \frac{1 + {x}^{2} - 2 {x}^{2}}{1 + {x}^{2}} ^ 2$

$\frac{\mathrm{dh}}{\mathrm{dx}} = - \frac{{x}^{2} - 1}{1 + {x}^{2}} ^ 2 < 0$ for $x > 1$

hence ${c}_{n + 1} < {c}_{n}$

Based on Leibniz' theorem then the series:

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} {b}_{n}$

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} {c}_{n}$

are both convergent, then by the squeeze theorem also:

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} {a}_{n}$

is convergent.