Re: Serie's problem

Thank you for your answer! Exactly what I was looking for. I prepeared some
Taylor series expansions with their sum functions for tomorrow.
In your message you write
"ln is the natural log (to the basee)." lol I really hope, that my posts
don't make me look so st*pid :) 10x again
Regards,
Dimo

acl wrote:
> Hello,
> here is how to do it: first of all, you need to know the taylor
> expansion
>
> ln(1-z)=-sum (z^n)/n,
> the sum from n=1 to infinity, and ln is the natural log (to the base
> e). This means that
> -ln(1+z)=sum ((-z)^n) /n.
> Next, rewrite your summand (the thing that is being summed) as
> ((-(1-x)/3)^n)/n,
> or, if we call z=(1-x)/3,
> ((-z)^n)/n,
> which makes your sum
> sum ((-z)^n) /n.
> But this is -ln(1+z)=-ln(1+(1-x)/3), which is the answer.
>
> To do this kind of thing you need to be able to spot immediately
> several standard series expansions (such as that for ln(1+x)) without
> thinking, which might be tricky in an exam! Good luck.

.

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