I'm trying to solve this problem:
Find the coefficients $a_k$ and the interval of convergence of the series: $f(x)=\frac{arctan(x)}{x+1}=\sum_{k=0}^\infty a_k(2x+1)^k$
So far what I've tried is to get the series representation of $arctan(x)$ centered in $(x-\frac{1}{2})$ like this:
$\frac{1}{1+x^2}=\frac{1}{\frac{5}{4}+(x^2-\frac{1}{4})}=\frac{1}{\frac{5}{4}}\frac{1}{1-(-\frac{4}{5}(x^2-\frac{1}{4}))}=\frac{4}{5}\sum_{k=0}^\infty (\frac{4}{5}(-(x^2-\frac{1}{4})))^k=\frac{4}{5}\sum_{k=0}^\infty(-1)^k\frac{4^k}{5^k}(x+\frac{1}{2})^k(x-\frac{1}{2})^k$
However I'm not sure how to proceed from here. Oh also I can easily get a series representation of $\frac{1}{1+x}$ of the form $\sum_{k=0}^\infty a_k(2x+1)^k$, but in that case I don't know how to incorporate the $arctan(x)$.
Best Answer
It is convenient to make a partial fraction decomposition since the linear terms in the denominator are then easier to expand.
Comment:
In (1) we can make the geometric series expansion and get a wanted representation in the form $\sum_{i=0}^\infty c_k(2x+1)^k$. We make some further transformations/simplifications to get rid of the imaginary unit.
In (2) we expand the terms of the series with the complex conjugate to obtain $\left(\frac{1}{2+i}\right)^{k+1}=\left(\frac{2-i}{(2+i)(2-i)}\right)^{k+1}=\frac{1}{5^{k+1}}(2-i)^{k+1}$ and similarly with the other part.
In (3) we make a binomial expansion and collect the terms conveniently in the next step.
In (4) we take even index $2l$ in the left-hand sum and odd index $2l+1$ in the right hand sum, since other indices do not contribute.
In (5) we use the binomial identity $\binom{p+1}{q}=\binom{p}{q}+\binom{p}{q-1}$.
Expansion of $\frac{1}{x+1}$ at $x=-\frac{1}{2}$ is not that hard. We have \begin{align*} \frac{1}{x+1}=\frac{1}{\frac{1}{2}\left(1+(2x+1)\right)}=2\sum_{k=0}^\infty(-1)^k(2x+1)^k\tag{8} \end{align*}
The radius of convergence is $\frac{1}{2}$ since the nearest singularity at the center $x=-\frac{1}{2}$ of the series expansion is $-1$.