Let $\displaystyle T_n = \sum_{i=1}^{n} \dfrac{(-1)^{i+1}}{2i-1}$, then show that $$\sum_{n=1}^{\infty} (T_n-T) = \dfrac{\pi}{8} -\dfrac{1}{4}$$ where $\displaystyle T = \lim_{n\rightarrow \infty} T_n$.
My approach for this was to define the function $\displaystyle F(x) = \sum_{n=1}\sum_{i=1} \dfrac{x^{i+n}}{2(i+n)-1}$ and figure out what $F(x)$ is. So far, I got $$\displaystyle 2F'(x)- \dfrac{F(x^2)}{x^2} = \dfrac{1}{(1+x)^2}.$$ I know $\arctan x$ shows up somewhere but that's about what I have. Any hints? I prefer hints to complete solutions.
Best Answer
$$T_n=\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}=\int_0^1\sum_{k=0}^{n-1}(-x^2)^k\,dx=\int_0^1\frac{1-(-x^2)^n}{1+x^2}\,dx;\\\sum_{n=1}^\infty(T_n-T)=\sum_{n=1}^\infty\int_0^1\frac{-(-x^2)^n}{1+x^2}\,dx=\int_0^1\frac{x^2\,dx}{(1+x^2)^2}.$$ Now write the integrand as $x\cdot x(1+x^2)^{-2}$ and integrate by parts.