# Find a function for this power series.

1. Aug 14, 2011

### Meatloaf4

1. The problem statement, all variables and given/known data
Perform a partial fractions expansion of 1/(n(n+1))=a/n + b/(n+1) in order to find a function that represents x^n/(n(n+1)).

2. Relevant equations

3. The attempt at a solution
so i broke up the partial fraction to 1/n -1(1+n). I integrate both to get ln(n) + ln(1-(-n)).. From there i am beyond lost

Its number six on the attached worksheet.

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2. Aug 14, 2011

### vela

Staff Emeritus
Can you state the entire problem as given? What you've written is confusing at best.

3. Aug 14, 2011

### Meatloaf4

Yea i apologize I attached a pdf with the question as its presented to me. Its number 6. If you like i can rewrite it if you don't feel like downloading the pdf.

4. Aug 14, 2011

### upsidedowntop

I'm not sure what you have been asked to do, but the following properties of power series might help you anyway.

$\frac{1}{1-x} = \sum x^n$

$f(x) = \sum a_n x^n \Rightarrow f'(x) = \sum na_n x^{n-1}$

$f(x) = \sum a_n x^n \Rightarrow \int^x f(x) = \sum \frac{a_n}{n+1} x^{n+1} + C$

I've ignored some assumptions about being within the radius of convergence and so forth.

5. Aug 14, 2011

### vela

Staff Emeritus
That helps. So far you found
$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$
So that means
$$\sum_{n=1}^\infty \frac{x^n}{n(n+1)} = \sum_{n=1}^\infty x^n\left(\frac{1}{n} - \frac{1}{n+1}\right) = \sum_{n=1}^\infty \left(\frac{x^n}{n} - \frac{x^n}{n+1}\right) = \sum_{n=1}^\infty \frac{x^n}{n} - \sum_{n=1}^\infty \frac{x^n}{n+1}$$
Now what you need to do is find what those two sums are equal to. Upsidedowntop has given you some good hints to help you do that.

6. Aug 14, 2011

### upsidedowntop

You need one more property of power series to answer these questions:
$f(x) = \sum a_n x^n \Rightarrow f(x)x^m = \sum a_n x^{n+m}$

I would feel bad about giving so many suggestions, except the 4 properties of power series I mentioned are properties you wouldn't be able to verify without some knowledge of sequences. ie, taking analysis. I think it's important to realize that they are not obvious, and that their proofs, though not terribly daunting, are outside the scope on an elementary calculus course.

Also, I checked the answers at the back. I agree with all of them except 3., which I think should be $\frac{x(x+1)}{(1-x)^3}$.

7. Aug 14, 2011

### Meatloaf4

Yeah I got to the portion vela is talking about, but im still lost from there. I'm assuming I have do something involving the power series representation of ln. The whole concept still kind of throws me for a loop. Any chance one of you could solve it for me just so I can see the steps. I literally tried for hours already, as well as looked at various things online for quite a while, i just don't get it.

8. Aug 14, 2011

### vela

Staff Emeritus
Here's an example of the kind of thing you need to do. Take the series x+2x2+3x3+...+nxn+.... If you pull a factor of x out, you'd get
$$x+2x^2+3x^3+\cdots = x(1+2x+3x^2+4x^3+\cdots)$$
Why would you want to do that? It's because the terms of the series in the parentheses now look like the derivative of xn, so you can say
\begin{align*}
x+2x^2+3x^3+\cdots &= x(1+2x+3x^2+4x^3+\cdots) \\
&= x\frac{d}{dx}(x+x^2+x^3+x^4+\cdots) \\
&= x\frac{d}{dx}(1+x+x^2+x^3+x^4+\cdots)
\end{align*}
You should recognize the series in the parentheses as the series for 1/(1-x), so now you have
\begin{align*}
x+2x^2+3x^3+\cdots &= x\frac{d}{dx}(1+x+x^2+x^3+x^4+\cdots) \\
&= x\frac{d}{dx}\left(\frac{1}{1-x}\right) \\
&= x\frac{1}{(1-x)^2} \\
&= \frac{x}{(1-x)^2}
\end{align*}

In your problem, you have xn/n. What does that look like it's the integral of? How about xn/(n+1)? What do you need to do to make it look like the result of an integration?

9. Aug 14, 2011

### SammyS

Staff Emeritus
For the $\displaystyle \sum_{n=1}^\infty\frac{x^n}{n}$ try the following:

Let $\displaystyle f(x)=\sum_{n=1}^\infty\frac{x^n}{n}$.

Now find the derivative of f(x).

The sum should now be a geometric series which can be evaluated.

Integrate that result to find f(x).

The other sum, $\displaystyle \sum_{n=1}^\infty\frac{x^n}{n+1}$ is a little more challenging but after a little playing around with it, you can get an expression with a sum similar to the one above.