How to rigorously prove that $\sum\limits_{n=1}^ \infty\left( \frac{1}{4n-1} - \frac{1}{4n}\right) =\frac{\ln(64)- \pi}{8}$ ?
My attempt
$$f_N(x):= \sum_{n=1}^ N \left(\frac{x^{4n-1}}{4n-1} - \frac{x^{4n}}{4n}\right)$$
$$f_N'(x) = \sum_{n=1}^ N( x^{4n-2}- x^{4n-1})= x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$$
I need to show that $x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$ converges uniformly to be able to interchange the derivative and the summation, but I don't think $f_N'$ converges uniformly because $$f_N'(x) = \frac{-x^2}{(1+x)(1+x^2)}\cdot (x^{4N+4}-1) $$ and $(x^{4N+4}-1)$ doesn't converge uniformly on $[0,1 )$.
Here I got stuck but for some reason it works, i.e., $\int_0 ^1 \frac{x^2}{(1+x)(1+x^2)}= \frac{\ln(64)- \pi}{8} $ so the derivative could be interchanged with the summation here, but how ?
If $x\in[-1,1]$, let$$f(x)=\sum_{n=1}^\infty\frac{x^{4n}}{4n(4n-1)};$$you want to compute $f(1)$. But, if $x\in(-1,1)$,\begin{align}f'(x)&=\sum_{n=1}^\infty\frac{x^{4n-1}}{4n-1}\\&=\frac{x^3}3+\frac{x^7}7+\frac{x^{11}}{11}+\cdots\\&=\int_0^xt^2+t^6+t^{10}+\cdots\,\mathrm dt\\&=\int_0^xt^2(1+t^4+t^8+\cdots)\,\mathrm dt\\&=\int_0^x\frac{t^2}{1-t^4}\,\mathrm dt\\&=\int_0^x-\frac1{2\left(1+t^2\right)}+\frac1{4 (1+t)}+\frac1{4(1-t)}\,\mathrm dt\\&=-\frac12\arctan(x)+\frac14\log(1+x)-\frac14\log(1-x).\end{align}Therefore\begin{align}f(1)&=\lim_{t\to1^-}f(t)-f(0)\quad\text{(since $f(0)=0$)}\\&=\lim_{t\to1^-}\int_0^tf'(x)\,\mathrm dx\\&=\lim_{t\to1^-}\int_0^t-\frac12\arctan(x)+\frac14\log(1+x)-\frac14\log(1-x)\,\mathrm dx\\&=\lim_{t\to1^-}\left[\frac{\log(1+x^2)}4-\frac x2\arctan(x)+\frac{1+x}4\log(1+x)+\frac{1-x}4\log(1-x)\right]_{x=0}^{x=t}\\&=\lim_{t\to1^-}\left(\frac{\log(1+t^2)}4-\frac t2\arctan(t)+\frac{1+t}4\log(1+t)+\frac{1-t}4\log(1-t)\right)\\&=\frac{\log(2)}4-\frac\pi8+\frac{\log(2)}2+0\\&=\frac{\log(64)-\pi}8.\end{align}