# Homework Help: Sum to Infinity

1. Apr 10, 2005

### crazedbeat

Hello!

Here is the problem I am attempting:

Sum to infinity:

$$\frac{x}{1*2} + \frac{x^2}{2*3} + \frac{x^3}{3*4} + ...$$

Here is what I get:

$$S = \frac{x^n}{(n)(n+1)}$$

$$\lim_{n \to \infty} \frac{x^n}{(n)(n+1)}}$$

Now what? Should I do partial fractions to split the equation?

2. Apr 10, 2005

### dextercioby

No,you should say the domain of "x"...You're interested in getting the value of the limit,right...?

Daniel.

3. Apr 10, 2005

### crazedbeat

I am trying to find the sum to infinity. Why should I do the domain of x?

4. Apr 10, 2005

### dextercioby

Because it involved a limit for which telling in which domain "x" takes values is ESSENTIAL...

Daniel.

5. Apr 10, 2005

### crazedbeat

I was not given any range of domain? Do I make up one. Essentially it can be anything...its in the numerator..

6. Apr 10, 2005

### dextercioby

It counts.If it's in $\left[0,1\right]$,the limit is 0.If it's in $\left(1,+\infty\right)$,then the limit is $+\infty$...

The tricky part is when "x" can be negative.

Daniel.

7. Apr 10, 2005

### crazedbeat

So if I take sum from negative infinity to positive infinity, all the places where x is raised to an odd power will get cancelled out. So only those which are left would be the one with the even power...which I take the sum of. However, how do I get the sum of the even power? i.e. how do I start this problem?

8. Apr 10, 2005

### crazedbeat

I was thinking one thing:

The power series is as such:

$$e^x = \frac{x^n}{n!}$$

Could I use this? All that's different about what I am doing is that there is also (n+1)! in the denominator.

So would I just do:

$$e^x * \sum_{n=0}^\infty \frac{1}{(n+1)!}$$

9. Apr 10, 2005

### quasar987

That would not help here. And you do not have factorials in your original S.

Also, I hope you know that what you have labeled S is just the nth term of the serie. This is different from the nth partial sum. Evaluating the limit you have writen will not give you the sum of the limit. All you can conclude from this limit is the range of points x where it does not converge (when lim does not equal 0), and the one where it MAY converge (when lim = 0).

10. Apr 10, 2005

### dextercioby

Nope,u can break it up.

Daniel.

11. Apr 10, 2005

### crazedbeat

But limit is only equal to 0 when x = 0....

Could someone kindly guide me in the right direction...I am really confused

12. Apr 10, 2005

### dextercioby

Here's the result.

$$\sum_{k=1}^\infty \frac{x^k}{k(k+1)}=\allowbreak \frac {1}{2}x\left( 2\frac{1-x}x-\frac {2}{x^2}\left[ \ln \left( 1-x\right) \right] \left( x-1\right) -\frac {1}{x-1}\left( -2x+2\right) \right)$$

I don't know how Maple did it,though...

Daniel.

13. Apr 10, 2005

### Hurkyl

Staff Emeritus
I think you should differentiate your series.

14. Apr 10, 2005

### dextercioby

If u do what Hurkyl said,u'll bump into this series

$$\sum_{k=1}^\infty \frac{x^{k-1}}{k+1}\allowbreak \allowbreak =\allowbreak \frac 1{x-1}\frac{1-x}x-\frac 1{x^2}\ln \left( 1-x\right)$$

Daniel.

P.S.I think you'll find useful the Taylor series for $\ln (1-x)$ around 0.

15. Apr 10, 2005

### crazedbeat

Why must one diffrentiate to this? I thought I was looking for the sum. Doesn't this do different all together?

16. Apr 10, 2005

### dextercioby

Nope,the result of the differentiation will be another series which can be computed more easily...

Daniel.

17. Apr 10, 2005

### crazedbeat

What does this series have to do with anything:

$$\sum_{k=1}^\infty \frac{x^{k-1}}{k+1}\allowbreak \allowbreak =\allowbreak \frac 1{x-1}\frac{1-x}x-\frac 1{x^2}\ln \left( 1-x\right)$$

at least this one has the correct left hand side:

$$\sum_{k=1}^\infty \frac{x^k}{k(k+1)}=\allowbreak \frac {1}{2}x\left( 2\frac{1-x}x-\frac {2}{x^2}\left[ \ln \left( 1-x\right) \right] \left( x-1\right) -\frac {1}{x-1}\left( -2x+2\right) \right)$$

But how is any of this to be evaluated?

18. Apr 10, 2005

### dextercioby

Differentiate the general term $\frac{x^{k}}{k(k+1)}$ wrt "x" and the u'll get the general term for the first series...

Daniel.

19. Apr 10, 2005

### crazedbeat

Perhaps I should explain first that our professor did not cover a lot of this stuff, I am an ECON major, and have been fiddling through a book which does not cover series. (Everything I know is through research online. And none of it makes sense.)

20. Apr 10, 2005

### crazedbeat

Oh alright that makes sense then.

$$\sum_{k=1}^\infty \frac{x^{k-1}}{k+1}\allowbreak \allowbreak =\allowbreak \frac 1{x-1}\frac{1-x}x-\frac 1{x^2}\ln \left( 1-x\right)$$

Now if I plug in one, I get a lot of illegal values, i.e. 0 in denominator and ln(0)......so I diffrentiate again for a better series?

21. Apr 10, 2005

### dextercioby

Sorry then for my persistance.I thought u wouldn't give up.

Daniel.

22. Apr 10, 2005

### dextercioby

No.That is the result.The sum of the series which,obviously depends on "x".That result should tell u however "the illegal values of <<x>>"...That "x" has lots of restrictions.

You should integrate now to get the original series.

Daniel.

23. Apr 10, 2005

### crazedbeat

No, no, I really appreciate your persistance. But could you just explain it a little bit to me? You have to understand I know very little about series-- only that I am suppose to add numbers. The first example I did, my friend did it with limits, so I thought that ALL could be done with limits. Now there is diffrentiation, which is just not makign sense :( and I feel like I am getting nowhere.

24. Apr 10, 2005

### dextercioby

Series,under certain conditions,especially of convergence,could be differentiated and integrated term by term.That's what it's done here.First u differentiate the original series to get another series which is simpler to evaluate.Then,u have to integrate back to get the original result.It's a very elegant method,but which only works,as i said,only if the series you get at each step are convergent...

Daniel.

25. Apr 10, 2005

### crazedbeat

Ooo. I see. But we haven't figured out the diffrentiated thing yet, so we shouldn't go back yet? If we go back and try and solve, we'd be nowhere, no?

Or have we figured it out and I completely missed it? Or do we integrate the simple thing and that'll give an answer? But what's the gurantee the simple thing integrated won't be complicated?