MacLaurin Series: Showing 1/n(n+1) = 1

[EEWannabe]

Homework Statement

Use the MacLaurin series for e^x and ln (1+x) to show that;

\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}...= 1

Homework Equations

e^{x}= 1 + x + \frac{x^{2}}{2!}+\frac{x^{3}}{3!}...

ln(1+x)= x - \frac{x^{2}}{2}+\frac{x^{3}}{3}...

The Attempt at a Solution

Well I'm not really sure how to go about this really, any sort of guidance would be great. Obviously it's in the form 1/n(n+1) but I'm not sure really :<

 

[micromass]

Not sure why you need the MacLaurin series of e^x and ln...

Observe that \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}. You can simplify the series with that...

 

[Chewy0087]

Hmmm that's where I'm getting confused.

I can see that you can expand as you said and use the method of differences to show that it's 1 but it says to use the expansions.

Also there are about 5/6 other questions on the sheet including

<br /> \frac{1}{1*2}-\frac{1}{2*3}+\frac{1}{3*4}-...= 1 <br />

So i guess there must be a trick for all of them using the expansions

edit: I am in the same class as this dude ;P we're talkin about it if you're confused

 

[EEWannabe]

yeah ^ what he said

 

[vela]

How about multiplying log(1+x) by x and then integrating?

 

[EEWannabe]

How about multiplying log(1+x) by x and then integrating?

well I get;

\frac{x^{3}}{3} - \frac{x^{4}}{4}+\frac{x^{5}}{5}-...

Which I can't see being very helpful...anyway you get a weird integral on the LHS of

\frac{x^{2}log(1+x)}{2}+integralof(\frac{x^{2}}{2(1+x)})

Maybe I'm being slow...but can you give me a bit more of a hint? ;s

 

[vela]

It would help if I integrated correctly.

Try integrating just log(1+x).

 

[EEWannabe]

It would help if I integrated correctly.

Try integrating just log(1+x).

Ah that's quite clever! ( i realized i integrated wrong before) Using it I was able to do the 2nd one, but not the first :S. (If you let x = -1 the LHS is messed up...)

 

[EEWannabe]

Also the next part is show that;

\frac{1}{1!}+\frac{4}{2!}+\frac{9}{3!}+\frac{16}{4!}...=2e

following on i tried messing round with integrals but came to nothing :( any advice would be great

 

[vela]

You could try taking the limit as x approaches -1⁺.

 

[vela]

You might have noticed by integrating, you managed to get a factor of n on the bottom of each term. What do you get when you differentiate e^x? Can you see how you might combine a sequence of differentiating and multiplying by powers of x to get the series you want?

 

[EEWannabe]

Yeah i worked it out about 20 minutes ago!

Thanks a lot for the help buddy, it's much obliged