# Series representation and other problems

1. Jun 14, 2004

### bard

find the series for sin(x)/x. I believe this would just mean dividing the series representation of sin(x) by x, therefore sin(x)/x=1-x^2/3!+x^4/5!-x^6/7!.....=sigma(x^2n/(2n+1)!)

how then would we find the radius of convergence and interval of convergence.

is the series n/sigma(1/k(k+2)) convergent or divergent. I believe the bottom is a telescoping series so it becomes 1/2 so then it becomes 2n and then lim(n-->infinity)=infinity and is therefire divergent. is this correct?

2. Jun 14, 2004

### AKG

Use the ratio test.

$$\lim _{n \rightarrow \infty} \left| \frac{x^{(2n+1)}/[2(n+1)+1]!}{x^{2n}/(2n+1)!} \right|$$

$$= \lim _{n \rightarrow \infty} \left| \frac{x^2}{(2n + 3)(2n + 2)} \right|$$

$$= 0$$

Therefore, it converges for all $x \in \mathbb {R}$; the radius of convergence, $R = \infty$ and the interval of convergence is $(-\infty , \infty )$.

What exactly is the series?

$$\sum _{k = 1} ^{\infty} \frac{1}{k(k+2)}$$

Is that it?

$$\forall k \in \mathbb {N} \ \ \ k(k + 2) > k^2$$

$$\therefore \frac{1}{k(k+2)} < \frac{1}{k^2}$$

By the p-series test (I beleive that's what it's called), we know that $\sum _{k=1} ^{\infty} \frac{1}{k^2}$ converges, and by the comparison test, we can conclude that the sum we're considering must also converge.

Last edited: Jun 14, 2004
3. Jun 15, 2004

### bard

$$\displaystyle{\lim_{n\to\infty}\frac{n}{\sum_{k=1}^n\frac{1}{k(k+2)}}}$$

4. Jun 15, 2004

### AKG

Okay, that's write, but it's not a series. It's just a limit, so you'd ask not whether it converges or diverges (I believe that applies to squences and series, and not to limits) but whethere or not it exists. As was shown, the series converges (assuming I didn't mess up anywhere), and I'll take your word that it converges to 1/2. Based on this, you're of course correct that:

$$\displaystyle{\lim_{n \rightarrow \infty}\frac{n}{\sum_{k=1} ^n\frac{1}{k(k+2)}}} = \infty$$

I.e. the limit does not exist.

EDIT: Notice, the answer to your first question should also be obvious because you know the function sin(x)/x has a finite value for all $x \neq 0$, so it obviously converges for those x, and approaches 1 as a limit as x approaches 0, so the series should converge there as well, I'm pretty sure.

Last edited: Jun 15, 2004
5. Jun 15, 2004

### bard

Thank You AKG, I need help with the following problem if you care to assist.

the integral of 2 sin(lnx)+ 1/x. u=ln(x). du=1/x so 2sin(u)+du=-2cos(u)+u=-2cos(lnx)+ln(x)

also the integral(from pi/2 to 0) of sinx+tanx.
I get this is really sin(x)+sin(x)/cos(x). so u=cos(x). du=-sinx dx so -du/u-du=-ln(cosx)-cos(x)?

but how do i evaulauate it at pi/2. thank you

6. Jun 15, 2004

### AKG

bard
Something's seriously wrong here. It would help if you used the LaTeX so I knew for sure what you meant, but I can certainly see a problem in "du = 1/x". If u = ln(x), then du = xdx. This should give you:

$$\int [2e^u \sin (u) + 1] du = 2 \int e^u \sin (u) du + u$$

Now, solve the integral:

$$I = \int e^u \sin (u) du$$

You'll find that you need to use integration by parts twice, and find that you'll have to subsitute an integral somewhere fore "I", then isolate "I" and divide by 2 I would assume (you'll figure it out when you do it) and solve for "I."

You're dealing with improper integrals. If you look at the graph of sin(x) + tan(x), or just tan(x) alone for that matter at $\pi /2$, you'll notice it's undefined. So, approach the problem as though you're solving for the definite integral from a to 0 as a approaches $\pi /2$. I would solve the integral by separating the sin(x) and tan(x). The antiderivative of sin(x) is cos(x), and the antiderivative of tan(x) is ln|sec(x)|. You'll probably memorize that eventually, but if you don't, you can figure it out yourself:

$$\int \tan (x) dx$$

$$= \int \frac{\sin (x)}{\cos (x)} dx$$

$$= \int \frac{u}{{\cos}^2 (x)} du$$

$$= \int \frac{u}{1 - u^2} du$$

$$= - \frac{1}{2} \int \frac{1}{v} dv$$

$$= - \frac{1}{2} \ln |v| + C$$

$$= \ln \frac{1}{\sqrt{|v|}} + C$$

$$= \ln \frac{1}{\sqrt{|1 - u^2|}} + C$$

$$= \ln \frac{1}{\sqrt{|{\cos}^2 (x)|}} + C$$

$$= \ln |\sec (x)| + C$$