Answers to the actual SOS Math stuff

Mathstud28 · **Posted:** Tue, 15 Jul 2008 05:02:17 UTC

Hello Moderators/Admins.

I cannot seem to send you a message! But it says to submit your answers, so here is some answers on this page

http://www.sosmath.com/calculus/series/ ... blems.html

Ok here is one of them for now

$$\sum_{n=1}^{\infty}\frac{\sin(n)}{n}$

We see that this a decreasing series of all positive terms so we may apply the Integral Test

$$\int_1^{\infty}\frac{\sin(x)}{x}\dx$

Well I do not know it exactly, but since

$$0\leq\int_1^{\infty}\frac{\sin(x)}{x}\dx\leq\int_0^{\infty}\frac{\sin(x)}{x}\dx$

and for the last integral, let

$\boxed{1}\quad$ $$J(\theta)=\int_0^{\infty}\frac{\sin(x)}{x}e^{-\theta{x}}\dx$

So then

$$J'(\theta)=\frac{d}{d\theta}\int_0^{\infty}\frac{\sin(x)}{x}e^{-\theta{x}}\dx=\int_0^{\infty}\frac{\partial}{\partial\theta}\bigg[\frac{\sin(x)}{x}e^{-\theta{x}}\bigg]\dx$

So

$$J'(\theta)=-\int_0^{\infty}\sin(x)e^{-\theta{x}}\dx$

Using integration by parts on that integral we get

$$J'(\theta)=-e^{-\theta{x}}\bigg[\frac{-(\cos(x)}{\theta^2+1}-\frac{\sin(x)\theta}{\theta^2+1}\bigg]\bigg|_{0}^{\infty}=\frac{-1}{\theta^2+1}$

So now we see looking at $\boxed{1}$

$$J(\infty)=\int_0^{\infty}\frac{\sin(x)}{x}e^{-\infty{x}}dx=0$

So $$\int_0^{\infty}J'(\theta)d\theta=\int_0^{\infty}\frac{-d\theta}{\theta^2+1}$

So we get

$J(\infty)-J(0)=\frac{-\pi}{2}$

Now seeing as stated that $J(\infty)=0$

We have

$-J(0)=\frac{-\pi}{2}\Rightarrow{J(0)=\frac{\pi}{2}}$

Now seeing once again by $\boxed{1}$

$$J(0)=\int_0^{\infty}\frac{\sin(x)}{x}e^{-(0)x}=\int_0^{\infty}\frac{\sin(x)}{x}$

We can conclude that

$\boxed{2}$ $$\int_0^{\infty}\frac{\sin(x)}{x}dx=\frac{\pi}{2}$

So now back to the pertinent matter, using $\boxed{2}$ we can see that

$$0\leq\int_1^{\infty}\frac{\sin(x)}{x}\dx\leq\frac{\pi}{2}$

$$\therefore\quad\int_1^{\infty}\frac{\sin(x)}{x}\dx\quad\text{is convergent}$

This finally enables us to say that $$\sum_{n=1}^{\infty}\frac{\sin(n)}{n}$ is convergent.

Mathstud28 · **Posted:** Tue, 15 Jul 2008 05:55:50 UTC

Here is another one

Question 14

$$\sum_{n=1}^{\infty}\frac{\cos\left(\ln(n)\right)}{n}$

Once again integral test

$$\int_1^{\infty}\frac{\cos\left(\ln(x)\right)}{x}=\sin\left(\ln(x)\right)\bigg|_1^{\infty}=\lim_{n\to\infty}\bigg[\sin\left(\ln(x)\right)\bigg|_1^{n}\bigg]=$ $$\lim_{n\to\infty}\sin\left(\ln(n)\right)-\sin\left(\ln(1)\right)=\lim_{n\to\infty}\sin\left(\ln(n)\right)$

This last limit is undefined, therefore the integral is undefined, and moreover the series is divergent.

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9632

Unfortunately the integral test is invalid for both series because the terms are not nonnegative.

For example, sin(4)/4 < 0 and cos(ln(6))/6 < 0.

Instead, you may want to consider Dirichlet's test.

Mathstud28 · **Posted:** Tue, 15 Jul 2008 07:40:18 UTC

Matt wrote:

Unfortunately the integral test is invalid for both series because the terms are not nonnegative.

For example, sin(4)/4 < 0 and cos(ln(6))/6 < 0.

Instead, you may want to consider Dirichlet's test.

Hahahahahahahahahahahahahahahahahaha, and series are supposed to be my speciality

. I cannot believe I missed that, well, at least people will see how to integrate $\frac{\sin(x)}{x}$ .

Wow, that is depressing that I did all that work, it didn't even occur to me.

Thanks Matt.

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9632

Mathstud28 wrote:

Wow, that is depressing that I did all that work, it didn't even occur to me.

I know... I considered not replying because I did not want to spoil the fun. But it happens.

Mathstud28 · **Posted:** Tue, 15 Jul 2008 07:52:36 UTC

Matt wrote:

Mathstud28 wrote:

Wow, that is depressing that I did all that work, it didn't even occur to me.

I know... I considered not replying because I did not want to spoil the fun. But it happens.

I am going to blame it on the lateness that I submitted it ::shifts eyes with insecurity::

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