<s>[INST] Showing that the power series converges at the endpoints of its interval of convergence The problem is the following:
Find the power series for $\frac{1}{2}\arctan^2(x)$ and show it converges in the endpoints of the interval of convergence.
My difficulty is in the convergence part.
The power series that I got:
$\displaystyle \frac{1}{2}\arctan^2(x) = \frac{x^2}{2} - \frac{x^4}{4}\left(\frac{1}{3} + 1\right) + \frac{x^6}{6}\left(\frac{1}{5} + \frac{1}{3} + 1\right) + ...$
The radius of convergence is $1$ $\implies$ $x\in(-1,1)$ it converges, but what about $x = 1$ or $x = -1$?
Since we have even powers, it is enough to show that it converges for $x = 1$.
The exercise gives a hint that says to use that:
$\displaystyle \left(\frac{1}{2n-1}+...+\frac{1}{5} + \frac{1}{3} + 1\right)<1+\frac{1}{2}\ln(2n-1)$
I was trying to find something convergent to compare using the innequality in the sum, but I didn't find anything.
Any help is appreciated.
 [/INST] Let $a_n=\frac1{2n+2}\sum_{k=0}^n\frac1{2k+1}.$
You will easily check that $(a_n)$ is decreasing, and the hint serves to prove it tends to $0.$
Then apply the alternating series test.
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