# How do you test for convergence for sum((-1)^n)*(sqrt(n))*(sin(1/n)) for n is 1 to infinity?

Jun 25, 2015

The series converges conditionally.

#### Explanation:

We have ${\sum}_{n = 1}^{+ \infty} {\left(- 1\right)}^{n} \cdot \sqrt{n} \cdot \sin \left(\frac{1}{n}\right)$.

Let's write it as ${\sum}_{n = 1}^{+ \infty} {\left(- 1\right)}^{n} \cdot {u}_{n}$,

where ${u}_{n} = \sqrt{n} \cdot \sin \left(\frac{1}{n}\right)$.

An alternating series is conditionally convergent if ${u}_{n}$ is decreasing (so ${u}_{n + 1} \le {u}_{n}$) for all $n \setminus > N$, $N \setminus \in \mathbb{N}$, and if ${\lim}_{n \to + \infty} {u}_{n} = 0$.

Let's prove ${u}_{n + 1} \le {u}_{n}$ for all $n$ big enough

Consider the function $f \left(x\right) = x \sin \left(\frac{1}{x}\right)$, $x \setminus > 0$.

Using the chain rule, we get :

$f ' \left(x\right) = \frac{1}{2 \sqrt{x}} \sin \left(\frac{1}{x}\right) - \frac{1}{x} ^ \left(\frac{3}{2}\right) \cos \left(\frac{1}{x}\right) = \frac{\frac{1}{2} x \sin \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right)}{x} ^ \left(\frac{3}{2}\right)$

Using the identity $\setminus \left\mid \sin \left(y\right) \right\mid \setminus \le q y$ for $y \setminus \ge q 0$, we get (putting $y = \frac{1}{x}$) :

$\frac{1}{2} x \sin \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right) \setminus \le q \frac{1}{2} - \cos \left(\frac{1}{x}\right)$ which goes to $- \frac{1}{2}$ when $x$ goes to infinity.

In particular, when $x$ is big enough, one has $f ' \setminus \le q 0$, i.e. $f$ is decreasing (so the sequence ${u}_{n}$ is also decreasing for $n$ big enough).

Now, let's calculate ${\lim}_{n \to + \infty} {u}_{n}$ :

${\lim}_{n \to + \infty} {u}_{n} = {\lim}_{n \to + \infty} \sqrt{n} \cdot \sin \left(\frac{1}{n}\right)$

You can't calculate the limit if ${u}_{n}$ stay like this, it would give you $' ' \left(+ \infty\right) ' ' \cdot 0$, which is undefined.

${\lim}_{n \to + \infty} \sqrt{n} \cdot \sin \left(\frac{1}{n}\right) = {\lim}_{n \to + \infty} \frac{\sin \left(\frac{1}{n}\right)}{\frac{1}{\sqrt{n}}}$

$= {\lim}_{n \to + \infty} \frac{\left(\sin \left(\frac{1}{n}\right)\right) '}{\left(\frac{1}{\sqrt{n}}\right) '}$ (L'Hospital's Rule)

$= {\lim}_{n \to + \infty} \frac{\cos \left(\frac{1}{n}\right) \cdot \left(- \frac{1}{n} ^ 2\right)}{- \frac{1}{2 \sqrt{{n}^{3}}}}$

$= {\lim}_{n \to + \infty} \cos \left(\frac{1}{n}\right) \cdot \left(\frac{1}{n} ^ 2\right) \cdot 2 \sqrt{{n}^{3}}$

$= {\lim}_{n \to + \infty} \frac{2 \sqrt{{n}^{3}}}{n} ^ 2 = {\lim}_{n \to + \infty} \frac{\sqrt{{n}^{3}}}{n} ^ 2$

$= {\lim}_{n \to + \infty} {x}^{- \frac{1}{2}} = {\lim}_{n \to + \infty} \frac{1}{\sqrt{x}} = 0$.

Therefore, the alternating series ${\sum}_{n = 1}^{+ \infty} {\left(- 1\right)}^{n} \cdot {u}_{n}$ converges conditionally.

Let's see if the series converges absolutely. It converges absolutely if the series of sequence ${u}_{n}$ converges.

We will use the comparison test for $n \ge 1$.

${u}_{n} = \sqrt{n} \cdot \sin \left(\frac{1}{n}\right) \ge \sin \left(\frac{1}{n}\right) = {v}_{n}$

We know that the series of sequence $\sin \left(\frac{1}{n}\right)$ diverges because, by comparison test,

${\lim}_{n \to + \infty} {v}_{n} / {h}_{n} = {\lim}_{n \to + \infty} \sin \frac{\frac{1}{n}}{\frac{1}{n}}$ (We use the harmonical series, which is divergent)

$= {\lim}_{n \to + \infty} \frac{\sin \left(\frac{1}{n}\right) '}{\left(\frac{1}{n}\right) '}$ (L'Hospital's Rule)

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

Since ${\lim}_{n \to + \infty} {v}_{n} / {h}_{n} \ne 0$ or ${\lim}_{n \to + \infty} {v}_{n} / {h}_{n} \ne + \infty$,

the series of sequence ${v}_{n}$ and the series of sequence ${h}_{n}$ are both convergent or divergent (by comparison test).
Since the series of sequence ${h}_{n}$ is divergent, the series of sequence ${v}_{n} = \sin \left(\frac{1}{n}\right)$ is divergent too.

Since ${u}_{n} \ge {v}_{n}$ and the series of sequence ${v}_{n}$ is divergent, the series of sequence ${u}_{n} = \sqrt{n} \cdot \sin \left(\frac{1}{n}\right)$ is divergent too (by comparison test).

Therefore, the alternating series ${\sum}_{n = 1}^{+ \infty} {\left(- 1\right)}^{n} \cdot {u}_{n}$ doesn't converge absolutely.

Apr 20, 2017

The series is convergent.

#### Explanation:

We know that

${\left(- 1\right)}^{n} \sqrt{n} \sin \left(\frac{1}{n}\right) = {\left(- 1\right)}^{n} \sqrt{n} \left(\frac{1}{n}\right) \left(\sin \frac{\frac{1}{n}}{\frac{1}{n}}\right) = {\left(- 1\right)}^{n} \frac{1}{\sqrt{n}} \left(\sin \frac{\frac{1}{n}}{\frac{1}{n}}\right)$

so as $n$ grows

${\left(- 1\right)}^{n} \sqrt{n} \sin \left(\frac{1}{n}\right) \approx {\left(- 1\right)}^{n} / \sqrt{n}$ and

${\sum}_{n} {\left(- 1\right)}^{n} / \sqrt{n}$ converges because is an alternating series and
$\left\mid {a}_{n} \right\mid = \frac{1}{\sqrt{n}}$ decreases monotonically to $0$

Finalizing, ${\sum}_{n} {\left(- 1\right)}^{n} \sqrt{n} \sin \left(\frac{1}{n}\right)$ converges.