A magnificent series for $\pi-333/106$

Probably not a very elegant method, as it makes a great use of a CAS. However it seems quite general for this kind of series. Inversely, it can be used to create similar results.

The terms of the series, which are a rational function of the index, can be decomposed into a sum of rational terms \begin{equation} u_k=\sum_{j=1}^n\frac{\lambda_j}{k+a_j} \end{equation} (we suppose the order of the poles is 1). When $\left|x\right|<1$, the series \begin{equation} f_j(x)=\sum_{k=0}^\infty \frac{x^k}{k+a_j}=x^{-a_j}\int_0^x \frac{t^{a_j}}{1-t}\,dt+\frac{1}{a_j} \end{equation} This can be verified by developping the $(1-t)^{-1}$ term in the integral. Then, the series \begin{align} S(x)&=\sum_{k=0}^\infty u_kx^k\\ &=\sum_{k=0}^\infty\sum_{j=1}^n\frac{\lambda_jx^k}{k+a_j}\\ &=\sum_{j=1}^n\lambda_j\left[x^{-a_j}\int_0^x \frac{t^{a_j}}{1-t}\,dt+\frac{1}{a_j}\right]\\ &=\sum_{j=1}^n\frac{\lambda_j}{a_j}+\int_0^x \frac{\sum_{j=1}^n\lambda_jx^{-a_j}t^{a_j}}{1-t}\,dt \end{align} The proposed series corresponds to $\lim_{x\to1^{-}}S(x)$. Due to the denominator in the integral, in order that this limit exists, the condition \begin{equation} \sum_{j=1}^n\lambda_j=0 \end{equation} must hold. Then \begin{equation} S(1)=\sum_{j=1}^n\frac{\lambda_j}{a_j}+\int_0^1 \frac{\sum_{j=1}^n\lambda_jt^{a_j}}{1-t}\,dt \end{equation} The remaining integral can be directly calculated.

In the proposed case, using a CAS, \begin{align} u_k&=\frac{48}{371}\frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)} \\ &=-{\frac {181203}{3799040\,k+21844480}}+{\frac {418643}{759808\,k+ 3229184}}-{\frac {293677}{759808\,k+2849280}}\\ &\,\quad+{\frac {743573}{3799040 \,k+12346880}}+{\frac {181203}{759808\,k+3988992}}-{\frac {1868267}{ 3799040\,k+18045440}}\\ &\,\quad-{\frac {56237}{759808\,k+2089472}}+{\frac {56237 }{3799040\,k+8547840}} \end{align} after some calculations, one obtains \begin{align} \sum_{j=1}^n\frac{\lambda_j}{a_j}&=\frac{7516928}{124151182155}\\ \sum_{j=1}^n\lambda_jt^{a_j}&=-{\frac {181203}{3799040}{t}^{{\frac{23}{4}}}}+{\frac {418643}{759808} {t}^{{\frac{17}{4}}}}-{\frac {293677}{759808}{t}^{{\frac{15}{4}}}}+{ \frac {743573}{3799040}{t}^{{\frac{13}{4}}}}\\ &\,\quad+{\frac {181203}{759808}{t }^{{\frac{21}{4}}}}-{\frac {1868267}{3799040}{t}^{{\frac{19}{4}}}}-{ \frac {56237\,{t}^{11/4}}{759808}}+{\frac {56237\,{t}^{9/4}}{3799040}}\\ &=\frac{1}{3799040} \left( 181203t+56237 \right)t^{9/4}\left( 1-\sqrt{t} \right)^5 \end{align} The above function vanishes at $t=1$, as expected. We have to evaluate \begin{align} S(1)&=\frac{7516928}{124151182155}+\frac{1}{3799040} \int_0^1 \frac{\left( 181203t+56237 \right)t^{9/4}\left( 1-\sqrt{t} \right)^5}{1-t}\,dt \\ &=\frac{7516928}{124151182155} +\frac{1}{949760}\int_0^1\frac{\left( 181203v^4+56237 \right)v^{12}\left( 1-v^2 \right)^5}{1+v^2}\,dv \end{align} To evaluate the integral, by devloping the numerator, we have to calculate terms as \begin{equation} I_n=\int_0^1\frac{v^{2n}}{1+v^2}\,dv \end{equation} A recurrence relation can be found easily: \begin{equation} I_n=\frac{1}{2n-1}-I_{n-1} \end{equation} from which we have (with $I_0=\pi/4$) \begin{equation} I_n=(-1)^{n-1}\sum_{p=0}^{n-1}\frac{(-1)^{p}}{2p+1}+(-1)^n\frac{\pi}{4} \end{equation} After (rather uninteresting) calculations, we get \begin{equation} \frac{1}{949760}\int_0^1\frac{\left( 181203v^4+56237 \right)v^{12}\left( 1-v^2 \right)^5}{1+v^2}\,dv=\pi-{\frac{780059253811}{248302364310}} \end{equation} Finally \begin{equation} S(1)=\pi-\frac{333}{106} \end{equation} as expected.

This series can be obtained with the same technique used in A series to prove $\frac{22}{7}-\pi>0$

Let us start from series $$\sum _{k=0}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=\pi-\frac{64}{21}$$

to obtain the following truncations:

$$\sum _{k=1}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=\pi-\frac{2176}{693}$$ $$\sum _{k=2}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=\pi-\frac{4288}{1365}$$ $$\sum _{k=3}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=\pi-\frac{45708032}{14549535}$$

The approximation we are interested in lies between two of these fractions.

$$ \frac{4288}{1365}< \frac{333}{106} < \frac{45708032}{14549535}$$

Therefore, a series for $\pi-\frac{333}{106}$ can be obtained as a mix of the series for $\pi-\frac{4288}{1365}$ and $\pi-\frac{45708032}{14549535}$.

From $$\pi-\frac{333}{106} = a(\pi-\frac{4288}{1365})+b(\pi-\frac{45708032}{14549535})$$

we obtain $$a=\frac{56237}{237440}$$ $$b=\frac{181203}{237440}$$

Finally,

$$\pi-\frac{333}{106}=\frac{56237}{237440}(\pi-\frac{4288}{1365})+\frac{181203}{237440}(\pi-\frac{45708032}{14549535})=$$ $$\frac{56237}{237440}\sum _{k=2}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}+$$ $$\frac{181203}{237440}\sum _{k=3}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=$$ $$\frac{56237}{237440}\sum _{k=0}^\infty \frac{960}{(4k+9)(4k+11)(4k+13)(4k+15)(4k+17)(4k+19)}+$$ $$\frac{181203}{237440}\sum _{k=0}^\infty \frac{960}{(4k+13)(4k+15)(4k+17)(4k+19)(4k+21)(4k+23)}=$$ $$\frac{48}{371} \sum_{k=0}^\infty \frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)}$$