# MacLaurin Series

## 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 :<

## Answers and Replies

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...

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

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

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

yeah ^ what he said

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

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 wierd 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
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It would help if I integrated correctly.

Try integrating just log(1+x).

It would help if I integrated correctly.

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...)

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

vela
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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?

Yeah i worked it out about 20 minutes ago!

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