How do you test for convergence for $\sum ({(- 1)}^{n}) \cdot (\sqrt{n}) \cdot (sin (\frac{1}{n}))$ $sum((-1)^n)(sqrt(n))(sin(1/n))$ for n is 1 to infinity?

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How do you test for convergence for $\sum ({(- 1)}^{n}) \cdot (\sqrt{n}) \cdot (sin (\frac{1}{n}))$ $sum((-1)^n)(sqrt(n))(sin(1/n))$ for n is 1 to infinity?

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Answer 1 · 2015-06-25T20:28:53

The series converges conditionally.

Explanation:

We have $+ \infty \sum n = 1 {(- 1)}^{n} \cdot \sqrt{n} \cdot sin (\frac{1}{n})$ $sum_(n=1)^(+oo)(-1)^n * sqrt(n) * sin(1/n)$ .

Let's write it as $+ \infty \sum n = 1 {(- 1)}^{n} \cdot u_{n}$ $sum_(n=1)^(+oo)(-1)^n * u_n$ ,

where $u_{n} = \sqrt{n} \cdot sin (\frac{1}{n})$ $u_n = sqrt(n) * sin(1/n)$ .

An alternating series is conditionally convergent if $u_{n}$ $u_n$ is decreasing (so $u_{n + 1} \leq u_{n}$ $u_(n+1)<=u_n$ ) for all $n > N$ $n \gt N$ , $N \in NN$ , and if $lim_(n->+oo)u_n = 0$ .

Let's prove $u_(n+1)<=u_n$ for all $n$ big enough

Consider the function $f(x) = xsin(1/x)$ , $x \gt 0$ .

Using the chain rule, we get :

$f'(x) = 1/(2 sqrt(x)) sin(1/x) - 1/x^(3/2) cos(1/x) = (1/2 x sin(1/x) - cos(1/x))/x^(3/2)$

Using the identity $\abs(sin(y)) \leq y$ for $y \geq 0$ , we get (putting $y = 1/x$ ) :

$1/2 x sin(1/x) - cos(1/x) \leq 1/2 - cos(1/x)$ which goes to $-1/2$ when $x$ goes to infinity.

In particular, when $x$ is big enough, one has $f' \leq 0$ , i.e. $f$ is decreasing (so the sequence $u_n$ is also decreasing for $n$ big enough).

Now, let's calculate $lim_(n->+oo) u_n$ :

$lim_(n->+oo) u_n = lim_(n->+oo) sqrt(n) * sin(1/n)$

You can't calculate the limit if $u_n$ stay like this, it would give you $''(+oo)''*0$ , which is undefined.

$lim_(n->+oo) sqrt(n) * sin(1/n) = lim_(n->+oo) (sin(1/n))/(1/sqrt(n))$

$=lim_(n->+oo) ((sin(1/n))')/((1/sqrt(n))')$ (L'Hospital's Rule)

$=lim_(n->+oo) (cos(1/n)*(-1/n^2))/(-1/(2sqrt(n^3)))$

$=lim_(n->+oo) cos(1/n)*(1/n^2)*2sqrt(n^3)$

$=lim_(n->+oo) (2sqrt(n^3))/n^2 = lim_(n->+oo)(sqrt(n^3))/n^2$

$=lim_(n->+oo) x^(-1/2)=lim_(n->+oo)1/sqrt(x)= 0$ .

Therefore, the alternating series $sum_(n=1)^(+oo)(-1)^n*u_n$ 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 $n>=1$ .

$u_n = sqrt(n)*sin(1/n) >= sin(1/n) = v_n$

We know that the series of sequence $sin(1/n)$ diverges because, by comparison test,

$lim_(n->+oo)v_n/h_n = lim_(n->+oo)sin(1/n)/(1/n)$ (We use the harmonical series, which is divergent)

$=lim_(n->+oo) (sin(1/n)')/((1/n)')$ (L'Hospital's Rule)

$=lim_(n->+oo) (cos(1/n)*(-1/n^2))/(-1/(n^2)) = lim_(n->+oo) cos(1/n) = 1$ .

Since $lim_(n->+oo)v_n/h_n != 0$ or $lim_(n->+oo)v_n/h_n != +oo$ ,

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 $v_n = sin(1/n)$ is divergent too.

Since $u_n >= v_n$ and the series of sequence $v_n$ is divergent, the series of sequence $u_n = sqrt(n) * sin(1/n)$ is divergent too (by comparison test).

Therefore, the alternating series $sum_(n=1)^(+oo)(-1)^n * u_n$ doesn't converge absolutely.

Answer 2 · 2017-04-20T23:59:48

The series is convergent.

Explanation:

We know that

$(-1)^n sqrt(n)sin(1/n)=(-1)^nsqrtn (1/n)(sin(1/n)/(1/n)) = (-1)^n1/sqrtn (sin(1/n)/(1/n))$

so as $n$ grows

$(-1)^n sqrt(n)sin(1/n) approx (-1)^n/sqrtn$ and

$sum_n (-1)^n/sqrtn$ converges because is an alternating series and
$abs(a_n)= 1/sqrtn$ decreases monotonically to $0$

Finalizing, $sum_n(-1)^n sqrt(n)sin(1/n)$ converges.

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