Series and integrals for inequalities and approximations to $\pi$

Fundamentals

Two beautiful expressions that relate $\pi$ to its convergents are Dalzell integral

$$\frac{22}{7}-\pi=\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx$$

and the following equivalent form of Lehmer series $$\pi-3=\sum_{k=1}^\infty \frac{24}{(4 k+1) (4 k+2) (4 k+4)}$$

(see http://matwbn.icm.edu.pl/ksiazki/aa/aa27/aa27121.pdf, page 139)

These are direct proofs that $\pi>3$ and $\frac{22}{7}>\pi$, because of the positiveness of the closed form under the summation and the integral, respectively.

Symmetry

Evaluating Lehmer series leads to the integral $$\pi-3=\int_0^1 \frac{4x^4(1-x)(2+x)}{(1+x)(1+x^2)}dx$$

Another relationship is given by the integral $$\pi-3=\int_{0}^{1} \frac{2 x (1-x)^2}{1+x^2} dx$$

and the series $$\pi-3=\sum_{k=0}^\infty \frac{24}{(4k+2)(4k+3)(4k+5)(4k+6)}$$

A series to prove $\frac{22}{7}-\pi>0$ is given by $$\sum_{k=1}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{22}{7}-\pi$$

which generalizes to $$\sum_{k=n}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\int_0^1 \frac{x^{4n}(1-x)^4}{1+x^2}dx$$

With this expression approximating fractions from the integrals in the RHS may be computed sequentially by adding the next term in the series.

For $n=0$ we have \begin{align} \frac{10}{3}-\pi &= \sum_{k=0}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)} \\ &= \int_0^1 \frac{(1-x)^4}{1+x^2}dx \\ \end{align}

and the same difference may be obtained from a series with three factors in the denominator and its corresponding integral

\begin{align} \frac{10}{3}-\pi &= \sum_{k=0}^\infty \frac{24}{(4k+4)(4k+6)(4k+7)} \\ &= \int_{0}^{1} \frac{4x^3(1-x)(1+2x)}{(1+x)(1+x^2)}dx \\ \end{align}

More convergents

For the third convergent, we have Lucas integral and a series \begin{align} \pi-\frac{333}{106} &= \int_0^1 \frac{x^5(1-x)^6(197+462x^2)}{530(1+x^2)}dx \\ &= \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)} \\ \end{align}

Finally, Lucas integral for the fourth convergent is $$\frac{355}{113}-\pi=\int_{0}^{1} \frac{x^8(1-x)^8(25+816x^2)}{3164(1+x^2)}dx$$ (see http://educ.jmu.edu/~lucassk/Papers/more%20on%20pi.pdf)

Q Is there a series for $\frac{355}{113}-\pi$?

• You seem to be really fascinated by this kind of problems. I enjoy to learn all of that from you. Thanks. – Claude Leibovici Feb 14 '16 at 7:38
• In what way do the others series correspond to their integrals? – Gerry Myerson Feb 14 '16 at 8:51
• @ClaudeLeibovici Thank you! They are fascinating and an answer leads to another question... – Jaume Oliver Lafont Feb 14 '16 at 8:57
• @GerryMyerson They evaluate to the same difference, I do not derive them from each other. – Jaume Oliver Lafont Feb 14 '16 at 8:58
• I have no proof but, $\displaystyle \dfrac{355}{113}-\pi=\sum_{k=0}^{+\infty}\dfrac{1489440+10321440k}{1243(4k+13)(4k+15)(4k+17)(4k+19)(4k+21)(4k+23)(4k+25)(4k+27)}$ – FDP Feb 14 '16 at 20:53

$\displaystyle \dfrac{355}{113}-\pi=\sum_{k=0}^{+\infty}\dfrac{1489440+10321440k}{1243(4k+13)(4‌​k+15)(4k+17)(4k+19)(4k+21)(4k+23)(4k+25)(4k+27)}$