# Given $T_n = \sum_{i=1}^{n} \frac{(-1)^{i+1}}{2i-1}$, show $\sum_{n=1}^{\infty} (T_n-T) = \frac{\pi}{8} -\frac{1}{4}$.

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.

• Hint: $T_n=\int_0^1\sum_{k=0}^{n-1}(-1)^k x^{2k}\,dx=\ldots$ so $T_n-T=\ldots$ and... Aug 12 '20 at 17:57
• Yeah I also thought of going that route and it gives us $\arctan(x)$ but I didn't see how that would be useful here Aug 12 '20 at 18:00

$$T_n=\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}=\int_0^1\sum_{k=0}^{n-1}(-x^2)^k\,dx=\int_0^1\frac{1-(-x^2)^n}{1+x^2}\,dx;\\\sum_{n=1}^\infty(T_n-T)=\sum_{n=1}^\infty\int_0^1\frac{-(-x^2)^n}{1+x^2}\,dx=\int_0^1\frac{x^2\,dx}{(1+x^2)^2}.$$ Now write the integrand as $$x\cdot x(1+x^2)^{-2}$$ and integrate by parts.
Note that $$L= \lim_{n\to \infty} T_n -T_n =\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2k-1}-\sum_{k=1}^{n}\frac{(-1)^{k-1}}{2k-1}=\sum_{k=1}^{\infty}\frac{(-1)^{k+n}}{2k-1+2n}=\sum_{k=1}^{\infty}(-1)^{k+n}\int_{0}^1 x^{2k+2n-2}dx =(-1)^{n}\int_0^1\left[\sum_{k=1}^{\infty} (-1)^{k-1}x^{2k+2n}\right]\frac{dx}{x^2}=(-1)^{n+1}\int_0^{1} \frac{x^{2n}}{1+x^2}dx$$ Therefore,$$\sum_{n\geq 1}L =\int_0^1\frac{1}{1+x^2}\left(\sum_{n\geq 1}(-1)^{n+1}x^{2n}\right)dx=\int_0^1\frac{x^2}{(1+x^2)^2}dx\underbrace {=}_{ x=\tan y} \int_0^{\frac{\pi}{4}}\sin^2ydy =\int_0^{\frac{\pi}{4}}\left(1-\cos^2y\right)dy=\frac{y}{2}-\frac{\sin2y }{4}\Biggr|_0^{\frac{\pi}{4}}=\frac{\pi}{8}-\frac{1}{4}$$