Floor value of infinite limit with sum If $\displaystyle l=\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{(\cos (2x))^{2r-1}}{2r-1}$ for $x\in(\cot^{-1}(2),\cot^{-1}(1))$. then $\lfloor l \rfloor $ is
What I tried:
put $\displaystyle \cos(2x)=t$ then $\displaystyle l=\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{t^{2r-1}}{2r-1}=\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\int x^{2r-2}dx$
$\displaystyle l=\lim_{n\rightarrow \infty}\int \sum^{n}_{r=1}x^{2r-2}dx=\lim_{n\rightarrow \infty}\int\frac{1-x^{2n}}{1-x^2}dx$
I do not know how to solve further.
Please assist.
 A: The Taylor series for $\log(1+ x)$ is
$$\log(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\,x^n$$
for $|x|<1$.  Hence, we see that 
$$\begin{align}
\log(1-x)-\log(1+x)&=-\sum_{n=1}^\infty \frac{1-(-1)^{n}}{n}\,x^n\\\\
&=-2\sum_{n=1}^\infty \frac{x^{2n-1}}{2n-1}\tag1
\end{align}$$
Substituting $\cos(2x)$ for $x$ in $(1)$ reveals
$$\log\left(\frac{1-\cos(2x)}{1+\cos(2x)}\right)=-2\sum_{n=1}^\infty \frac{\cos^{2n-1}(2x)}{2n-1}\tag2$$
Dividing both sides of $(2)$ by $-2$ yields
$$\log(\cot(x))=\sum_{n=1}^\infty \frac{\cos^{2n-1}(2x)}{2n-1}\tag3$$
For $x\in (\text{arccot}(2), \text{arccot}(1))$, then $\cot(x)\in (1,2)$ and therefore $\log(\cot(x))\in (0,\log(2))$.  Applying the floor function to $\log(\cot(x))$ yields $0$. 

If one proceeds as in the OP, we first see that $t=\cos(2x)\in (0,3/5)$ for $x\in (\text{arccot}(2),\text{arccot}(1)) $.  Then, we have
$$\begin{align}
\sum_{r=1}^n \frac{t^{2r-1}}{2r-1}&=\int_0^t \sum_{r=1}^n s^{2r-2}\,ds\\\\
&=\int_0^t\frac{1-s^{2n}}{1-s^2}\,ds\\\\
&=\log\left(\sqrt{\frac{1+t}{1-t}}\right)-\int_0^t\frac{s^{2n}}{1-s^2}\,ds\tag4
\end{align}$$
As $n\to\infty$ the value of the integral on the right-hand side of $(4)$ approaches $0$.  Therefore, 
$$\lim_{n\to\infty}\sum_{r=1}^n \frac{t^{2r-1}}{2r-1}=\log\left(\sqrt{\frac{1+t}{1-t}}\right)\tag5$$
With $t=\cos(2x)$ in $(5)$ we find
$$\lim_{n\to\infty}\sum_{r=1}^n \frac{\cos^{2r-1}(2x)}{2r-1}=\log\left(\sqrt{\frac{1+\cos(2x)}{1-\cos(2x)}}\right)=\log(\cot(x))\tag6$$
which agrees with $(3)$. And we can conclude as in the previous development.
