Finding the sum of the series $\sum_{n=0}^n \frac{x^{4n}}{4n+3}$ $$\sum_{n=0}^n \frac{x^{4n}}{4n+3}$$
Now we're given a hint to turn this into $x^{-3}\int g(t)dt$
I've gone ahead and done the following:
$$x^{-3}\sum_{n=0}^n \frac{x^{4n+3}}{4n+3}$$
And this turns into:
$$x^{-3}\sum_{n=0}^n \left(\int_0^x{t^{4n+2}}\right)$$
Now we know for a power series:
$$\sum_{n=0}^n q^{n}= \frac{1}{1-q}$$
And I need to turn the term from two lines ago into this form. 
$t^{4n+2}=t^{4n}{t^2}=\left((t^4)^{n}*t^2\right)$
Now is my thought process correct here?
$$x^{-3}\sum_{n=0}^n \left(\int_0^x{t^{4n+2}}\right)=x^{-3}t^2\sum_{n=0}^n \left(\int_0^x{t^{4n}}\right)=x^{-3}\int\left(\frac{t^2}{1-t^4}\right)$$
(And of course after this comes the fraction decomposition and the integral.)
 A: Fine so far,
$$\sum_{n\geq 0}\frac{x^{4n+3}}{4n+3} = \int_{0}^{x}\frac{t^2}{1-t^4}\,dt = \frac{\text{arctanh}(x)-\arctan(x)}{2}$$
leads to:
$$ \sum_{n\geq 0}\frac{x^{4n}}{4n+3} = \frac{\text{arctanh}(x)-\arctan(x)}{2x^3}$$
for any $x\in(-1,1)$.
A: From the OP

Now is my thought process correct here?$$x^{-3}\sum_{n=0}^n \left(\int_0^x{t^{4n+2}}\right)=x^{-3}t^2\sum_{n=0}^n \left(\int_0^x{t^{4n}}\right)=x^{-3}\int\left(\frac{t^2}{1-t^4}\right)
$$

Not quite.  There is a conceptual error when the dummy variable was pulled outside the integral in the middle expression. It was inserted tacitly back under the integral (another conceptual error) to arrive at a correct result. 
To proceed correctly, we write
$$\begin{align}
x^{-3}\sum_{n=0}^\infty \int_0^x t^{4n+2}\,dt&=x^{-3}\lim_{N\to \infty}\int_0^x \sum_{n=0}^N t^{4n+2}\,dt\\\\
&=x^{-3}\lim_{N\to \infty}\int_0^x t^2\frac{1-t^{4N+4}}{1-t^4}\,dt\\\\
&=x^{-3}\int_0^x \frac{t^2}{1-t^4}\,dt
\end{align}$$
where the interchange of the limit with the integral is justified since the integrand converges uniformly on $x\in [-r,r]$ for all $0<r<1$
A: $$x^{-3}\sum_{i=0}^n \left(\int_0^x{t^{4n+2}} dt\right)=x^{-3} \left(\int_0^x t^2 \sum_{i=0}^n{t^{4n}} dt\right) = x^{-3} \left(\int_0^x t^2 \frac{1-t^{4n+4}}{1-t^4} dt\right) = x^{-3} \left(\int_0^x t^2 \frac{1}{1-t^4} dt\right)-x^{-3} \left(\int_0^x t^{4n+6} \frac{1}{1-t^4} dt\right).$$
The first integral is $$\dfrac{\ln\left(\left|x+1\right|\right)-2\arctan\left(x\right)-\ln\left(\left|x-1\right|\right)}{4},$$ which is obtained by partial fractions and the second is also solved by partial fractions of $1+t^2$, $1-t$, and $1+t$ to get
$$\dfrac{\ln\left(\left|x+1\right|\right)}{4}- \dfrac{\arctan\left(x\right)}{2}-\sum_{i=0}^n\dfrac{x^{4i+3}}{4i+3}-\dfrac{\ln\left(\left|x-1\right|\right)}{4}.$$
Thus, we will not get much for finite $n$ since subtracting leads to original problem. However, for $n$ infinite, we do get the first term as answer. I do not see that we can solve for finite $n$ using this approach.
