Re: Serie's problem

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.

Dimo Pepelyashev wrote:
> No I am not! It is the inversed way. I use Taylor axpansion (or maybe it's
> better to call it collapsion) to find the root function (it is my answer).
> But It looks very hard or I am missing something? Can the calc do the job?
> Sorry if I am asking simple questions, but I have an exam tomorrow and I
> think I took a big portion of math these days :)
> Thanks for the patience!

.

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