Let $T_n=\sum\limits_{i=1}^n\frac{(-1)^{i+1}}{2i-1}$ and $T=\lim T_n$, show that $\sum\limits_{n=1}^\infty(T_n-T)=\frac{\pi-2}8$ 
Let $T_n=\sum\limits_{i=1}^{n}\frac{(-1)^{i+1}}{2i-1}$ and $T=\lim\limits_{n \to\infty}T_n$, show that $\sum\limits_{n=1}^\infty(T_n-T)=\frac{\pi}{8}-\frac{1}{4}$.

Attempt: 
I know that $Im(\frac{1}{2}log(\frac{1+i}{1-i}))=T$, but I have no idea how to evaluate $Im(\frac{1}{2}log(\frac{1+i}{1-i}))$
 A: I expand my comment into an answer with rigorous justification of all the steps.

We have $$\frac{1}{1 + x^{2}} = 1 - x^{2} + x^{4} - \cdots + (-1)^{n}\frac{x^{2n}}{1 + x^{2n}}\tag{1}$$ Integrating with respect to $x$ over interval $[0, 1]$ we get $$\int_{0}^{1}\frac{dx}{1 + x^{2}} = T_{n} + (-1)^{n}R_{n}\tag{2}$$ where $$R_{n} = \int_{0}^{1}\frac{x^{2n}}{1 + x^{2}}\,dx\tag{3}$$ Clearly we can see from $(3)$ that $$0 \leq R_{n} \leq \int_{0}^{1}x^{2n}\,dx = \frac{1}{2n + 1}$$ and hence by Squeeze theorem we have $\lim_{n\to\infty}R_{n} = 0$ and hence taking limit of $(2)$ when $n\to\infty$ we get $$\int_{0}^{1}\frac{dx}{1 + x^{2}} = \lim_{n \to \infty}T_{n}$$ so that $$T = \int_{0}^{1}\frac{dx}{1 + x^{2}}$$ Therefore from $(2)$ we get $$T_{n} - T = (-1)^{n + 1}\int_{0}^{1}\frac{x^{2n}}{1 + x^{2}}\,dx\tag{4}$$ and hence $$\sum_{k = 1}^{n}(T_{k} - T) = \int_{0}^{1}\frac{1}{1 + x^{2}}\sum_{k = 1}^{n}(-1)^{k + 1}x^{2k}\,dx$$ or $$\sum_{k = 1}^{n}(T_{k} - T) = \int_{0}^{1}\frac{x^{2} + (-1)^{n + 1}x^{2n + 2}}{(1 + x^{2})^{2}}\,dx$$ or $$\sum_{k = 1}^{n}(T_{k} - T) = \int_{0}^{1}\frac{x^{2}}{(1 + x^{2})^{2}}\,dx + (-1)^{n + 1}P_{n}\tag{5}$$ where $$P_{n} = \int_{0}^{1}\frac{x^{2n + 2}}{(1 + x^{2})^{2}}\,dx$$ And like before we can see that $$0 \leq P_{n} \leq \int_{0}^{1}x^{2n + 2}\,dx = \frac{1}{2n + 3}$$ and hence by Squeeze we see that $P_{n} \to 0$ as $n \to \infty$. Hence taking limit of $(5)$ as $n \to \infty$ we get $$\sum_{k = 1}^{\infty}(T_{k} - T) = \int_{0}^{1}\frac{x^{2}}{(1 + x^{2})^{2}}\,dx = \int_{0}^{\pi/4}\sin^{2}t\,dt = \frac{\pi - 2}{8}$$ where we use the substitution $x = \tan t$ in the last step.
A: You are over-complicating things. It is well-known that $T=\sum_{k\geq 1}\frac{(-1)^{k+1}}{2k-1}=\frac{\pi}{4}$, and by summation by parts:
$$\sum_{n=1}^{N}(T_n-T) = N(T_N-T)-\sum_{n=1}^{N-1}n (T_{n+1}-T_n)=\sum_{k\geq N}N\frac{(-1)^{k+1}}{2k+1}+\sum_{n=1}^{N-1}n\frac{(-1)^{n+1}}{2n+1} $$
can be re-written as
$$ \sum_{n\geq 1}\min(n,N)\frac{(-1)^{n+1}}{2n+1}\stackrel{N\to+\infty}{\longrightarrow}\sum_{n\geq 1}'\frac{n}{2n+1}(-1)^{n+1} $$
by the dominated convergence theorem, where $\sum'$ stands for a Cesàro sum. Since $\sum_{n\geq 1}'\frac{1}{2}(-1)^{n+1}=\frac{1}{4}$, your series equals:
$$ \frac{1}{4}-\frac{1}{2}\sum_{n\geq 1}\frac{(-1)^{n+1}}{2n+1}=\frac{1}{4}-\frac{4-\pi}{8}=\color{red}{\frac{\pi-2}{8}}.$$
