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?
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.