MacLaurin Series

  • Thread starter EEWannabe
  • Start date
  • #1
32
0

Homework Statement


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

[tex] \frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}...= 1 [/tex]

Homework Equations



[tex] e^{x}= 1 + x + \frac{x^{2}}{2!}+\frac{x^{3}}{3!}....[/tex]

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


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 :<
 

Answers and Replies

  • #2
22,129
3,297
Not sure why you need the MacLaurin series of e^x and ln...

Observe that [tex]\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} [/tex]. You can simplify the series with that...
 
  • #3
370
0
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

[tex]
\frac{1}{1*2}-\frac{1}{2*3}+\frac{1}{3*4}-...= 1
[/tex]

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

edit: im in the same class as this dude ;P we're talkin about it if you're confused
 
  • #4
32
0
yeah ^ what he said
 
  • #5
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,094
1,674
How about multiplying log(1+x) by x and then integrating?
 
  • #6
32
0
How about multiplying log(1+x) by x and then integrating?

well I get;

[tex] \frac{x^{3}}{3} - \frac{x^{4}}{4}+\frac{x^{5}}{5}-....[/tex]

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

[tex] \frac{x^{2}log(1+x)}{2}+integralof(\frac{x^{2}}{2(1+x)}) [/tex]

Maybe i'm being slow...but can you give me a bit more of a hint? ;s
 
  • #7
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,094
1,674
It would help if I integrated correctly. :wink:

Try integrating just log(1+x).
 
  • #8
32
0
It would help if I integrated correctly. :wink:

Try integrating just log(1+x).

Ah that's quite clever! ( i realised 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...)
 
Last edited:
  • #9
32
0
Also the next part is show that;

[tex] \frac{1}{1!}+\frac{4}{2!}+\frac{9}{3!}+\frac{16}{4!}...=2e[/tex]

following on i tried messing round with integrals but came to nothing :( any advice would be great
 
  • #10
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,094
1,674
You could try taking the limit as x approaches -1+.
 
  • #11
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,094
1,674
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 ex? Can you see how you might combine a sequence of differentiating and multiplying by powers of x to get the series you want?
 
  • #12
32
0
Yeah i worked it out about 20 minutes ago!

Thanks alot for the help buddy, it's much obliged
 

Related Threads on MacLaurin Series

  • Last Post
Replies
9
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
8K
  • Last Post
Replies
8
Views
906
  • Last Post
Replies
2
Views
758
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
16
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
3
Views
957
Top