The first part of my question (for my homework) states "Find the Taylor series about $x=0$ of the function $f(x)=\frac{1}{(1-x)^2}$ "...
I have found the taylor series of $\frac{1}{(1-x)^2}$ to be $$\sum_{n=0}^\infty\dfrac{\dfrac{1}{(1-a)^2}^n{(x-a)^n}}{n!}$$ and the Maclaurin series to be $$\sum_{n=0}^\infty(n+1)x^n$$
I have a couple of questions...
- Am I correct in starting my bound as $n=0$? I wasn't sure because some examples have $n=1$. Maybe that's just if the function is undefined for $n=0$?
- I have found both the taylor series and the maclaurin series as the question asks for the "Taylor series" about $x=0$ which i thought was specifically called a Maclaurin series? But they are different series, so... I am asked to then find the radius and interval of convergence of the answer but I am really confused as to whether I am meant to use the Taylor series version or the Maclaurin series version to do this. Can someone clarify which series out of the ones I found to use for this part?
- The last half of the question states "Hence or otherwise determine the sum of the series summed from $n=0$ to infinity of $\frac{(n+1)}{2^n}$... Does that part have anything to do with either of the series I found for the first half? Or is it just a new part of the question and I should apply the same techniques to find the sum of that series as any other and just ignore the first half of the question to complete this part?
Your last question: on your second attempt you successfully computed the Maclaurin series as $$\frac1{(1-x)^2}=\sum_{n=0}^\infty(n+1)x^n.\tag{$*$}$$ You are enjoined to calculate $$\sum_{n=0}^\infty\frac{n+1}{2^n}.$$ This looks very similar to the RHS of $(*)$. Can you obtain it from the RHS of $(*)$ by making a substitution?