Find a function for this power series.

[Meatloaf4]

Homework Statement

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)).

Homework Equations

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.

 

[vela]

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

 

[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.

 

[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.

 

[vela]

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.

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.

 

[upsidedowntop]

Two more comments:

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 $\frac{x(x+1)}{(1-x)^3}$.

 

[Meatloaf4]

Yeah I got to the portion vela is talking about, but I am 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.

 

[vela]

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?

 

[SammyS]

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.