5

Series for sums $\pi+\dfrac{p_n}{q_n}$

Lehmer's interesting series relating $\pi$ to its early convergents $3$, $\dfrac{22}{7}$ and $\dfrac{355}{113}$ may be written as follows.

$$\begin{align} \sum_{n=1}^\infty \dfrac{n2^n}{\dbinom{2 n}{n}}&=\pi+3\tag{A.1}\\ \dfrac{2}{7}\sum_{n=1}^\infty \dfrac{n^22^n}{\dbinom{2 n}{n}}&=\pi+\dfrac{22}{7}\tag{A.2}\\ \dfrac{2}{35}\sum_{n=1}^\infty \dfrac{n^32^n}{\dbinom{2 n}{n}}&=\pi+\dfrac{22}{7}\tag{A.3}\\ \dfrac{1}{113}\sum_{n=1}^\infty \dfrac{n^42^n}{\dbinom{2 n}{n}}&=\pi+\frac{355}{113}\tag{A.4}\\ \end{align}$$

Series for differences $\pi-\dfrac{p_n}{q_n}$ or $\dfrac{p_n}{q_n}-\pi$

Series that prove the sign of the error when approximating $\pi$ by its convergents include:

$$\begin{align} \pi-3&=4·24\sum_{k=0}^\infty \frac{(4k+1)!}{(4k+6)!}(k+1)\tag{B.1}\\ \frac{22}{7}-\pi&=4^2·240\sum_{k=0}^\infty \frac{(4k+3)!}{(4k+11)!}(k+1)(k+2)\tag{B.2}\\ \frac{22}{7}-\pi&=4^3·285120\sum_{k=0}^\infty \frac{(4k+1)!}{(4k+14)!}(k+1)(k+2)(k+3)\tag{B.3}\\ \end{align}$$

A series to prove $\frac{22}{7}-\pi>0$

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

Changing the initial value for summations leads to results as in A.1, A.2 and A.3.

$$\begin{align} \pi+3&=4·24\sum_{k=-1}^\infty \frac{(4k+1)!}{(4k+6)!}(k+1)\tag{C.1}\\ \frac{22}{7}+\pi&=-4^2·240\sum_{k=-3}^\infty \frac{(4k+3)!}{(4k+11)!}(k+1)(k+2)\tag{C.2}\\ \frac{22}{7}+\pi&=-4^3·285120\sum_{k=-3}^\infty \frac{(4k+1)!}{(4k+14)!}(k+1)(k+2)(k+3)\tag{C.3}\\ \end{align}$$

Let us assume there is a series for the fourth convergent

$$\sum_{k=0}^\infty \frac{q}{\Pi_{i}(4k+d_i)}=\frac{355}{113}-\pi\tag{B.4}$$

for some positive rational $q$ and positive integers $d_i$.

Question: Is there a technique to obtain the difference series (B.1, B.2, B.3) from their sum counterparts (A.1, A.2, A.3) that allows to determine B.4 from A.4?

Jaume Oliver Lafont
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  • A failed attempt using A type series only is $$\frac{1}{113}\sum_{k=0}^\infty \dfrac{(n^4-2·113(2n-3))2^n}{\dbinom{2n}{n}} = \frac{355}{113}-\pi$$ because it has positive and negative terms. – Jaume Oliver Lafont May 20 '17 at 04:14
  • Although $4$ is not a convergent of $\pi$, $1$ is a convergent of $\dfrac{\pi}{4}$, so we may want to add $$\begin{align} 2\sum_{n=1}^\infty\frac{2^n}{\dbinom{2n}{n}}&=4+\pi\tag{A.0}\ \sum_{k=0}^\infty \frac{8}{(4k+3)(4k+5)}&=4-\pi\tag{B.0}\ \end{align}$$

    although this would make factor $(k+1)$ come up at $B.0$...

     – Jaume Oliver Lafont May 20 '17 at 04:40
  • $1,1,2,3...$ $\frac{355}{113}-\pi$ might have $5$ factors instead of $4$ $$(k+1)(k+2)(k+3)(k+4)(k+5)$$ – Jaume Oliver Lafont May 20 '17 at 05:00

1 Answers1

1

Since the second series for $\frac{22}{7}-\pi$ comes from

$$\frac{1}{28}\int_0^1 \frac{x(1-x)^8(2+7x+2x^2)}{1+x^2} dx = \frac{22}{7}-\pi$$

the following integral could be a starting point.

$$\frac{1}{99440} \int_0^1 \frac{x^{11}(1-x)^{12}( 858416(1+x^2)+6215x)}{1+x^2}dx = \frac{355}{113}-\pi$$

Maybe smaller coefficients would help such as in

$$\frac{1}{3164}\int_0^1 \frac{x^9(1-x)^8(791x+50(1+x^2))}{1+x^2} dx = \frac{355}{113} -\pi,$$

but the series obtained applying the technique in https://math.stackexchange.com/a/1657416/134791 is

$$\frac{4561920}{113} \sum_{k=0}^\infty \frac{ (81 k^2 + 648 k + 1145) (4 k + 16)}{ (4 k + 10)_{13} } = \frac{355}{113} - \pi$$

not as simple as the ones in the question.

Jaume Oliver Lafont
  • 5,212