0
$\begingroup$

$$\sum_{n=1}^{\infty} \frac{(-1)^n \cos^2(2n)}{\sqrt{n}}$$

In case of conditional convergence, I suppose series is convergent (Dirichlet's test). But I don't know how to prove that $\sum_{n=1}^{m} \cos^2(2n)$ is bounded. In case of absolute convergence I don't have good idea. Could you give me any tips?

$\endgroup$
3
  • $\begingroup$ How the plan of letting us doing all your homeworks is going on? $\endgroup$ Commented Nov 3, 2016 at 23:54
  • 1
    $\begingroup$ Tip: $\sum_{n=1}^{+\infty}\frac{(-1)^n\cdot\frac{1}{2}}{\sqrt{n}}$ is clearly convergent, and $\frac{1}{2}-\cos^2(2n)$ can be developped as a Fourier cosine series. By Dirichlet's test, $$\sum_{n=1}^{+\infty}\frac{(-1)^n \cos(mn)}{\sqrt{n}}$$ is convergent for every $m\in\mathbb{N}$. $\endgroup$ Commented Nov 3, 2016 at 23:56
  • $\begingroup$ By similar reasons (summation by parts), the original series is not absolutely convergent, $$\sum_{n=1}^{N}\frac{\cos^2(2n)}{\sqrt{n}}\approx\sqrt{N}.$$ $\endgroup$ Commented Nov 4, 2016 at 0:01

2 Answers 2

1
$\begingroup$

To use Dirichlet's Test, you need to prove that \begin{align} \sum^N_{n=1} (-1)^n\cos^2(2n) \end{align} is bounded. Observe \begin{align} \sum^{N}_{n=1} (-1)^n\cos^2(2n) =&\ \sum^N_{n=1}(-1)^n\left(\frac{1+\cos(4n)}{2} \right) \\ =&\ \sum^N_{n=1}(-1)^n\frac{1}{2} +\frac{1}{2}\operatorname{Re}\sum^N_{n=1} (-1)^ne^{i4n}\\ =&\ \sum^N_{n=1}(-1)^n\frac{1}{2}+ \frac{1}{2}\operatorname{Re} \frac{-e^{i4}-(-1)^{N+1}e^{i4(N+1)}}{1+e^{i4}} \end{align} which is bounded since \begin{align} \sum^N_{n=1}(-1)^n\frac{1}{2} = \begin{cases} -\frac{1}{2} & \text{ if } N \text{ is odd}\\ 0 &\ \text{ if } N \text{ is even} \end{cases} \end{align} and \begin{align} \left|\frac{e^{i4}+(-1)^{N+1}e^{i4(N+1)}}{1+e^{i4}} \right|< \frac{1}{|\cos 2|}. \end{align}

For the absolute convergence part note that we are considering the series \begin{align} \sum^\infty_{n=1} \frac{1+\cos(4n)}{2\sqrt{n}}. \end{align}

Again applying Dirichlet's Test, we see that \begin{align} \sum^\infty_{n=1} \frac{\cos(4n)}{2\sqrt{n}}<\infty \end{align} which means \begin{align} \sum^\infty_{n=1} \frac{1+\cos(4n)}{2\sqrt{n}} = \infty. \end{align}

$\endgroup$
1
  • $\begingroup$ @Jack'swastedlife You are right. I have made the correction. $\endgroup$ Commented Nov 7, 2016 at 22:14
0
$\begingroup$

One way to see $\sum_{n=1}^{\infty}\cos^2(2n)/\sqrt n = \infty$ is to notice that $\cos^2(2x) + \cos^2(2(x+1))$ is periodic (with period $\pi/2$) and is never $0.$ It follows that there is $c>0$ such that $\cos^2(2x) + \cos^2(2(x+1)) > c$ for all $x.$ Thus for all $n,$

$$\frac{\cos^2(2n)}{\sqrt n} + \frac{\cos^2(2(n+1))}{\sqrt {n+1}} \ge \frac{\cos^2(2n)}{\sqrt {n+1}} + \frac{\cos^2(2(n+1))}{\sqrt {n+1}}> \frac{c}{\sqrt {n+1}}.$$

Since $\sum c/\sqrt {n+1} = \infty,$ we have shown divergence.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .