The approximation
$$\pi\approx\frac{22}{7}=3+\frac{1}{7}$$
suggests that the closest integer to $\frac{1}{\left(\pi-3\right)}$ is $7$.
However, $$ \frac{1}{\left(\pi-3\right)^2}\approx49.879 $$ is closer to 50 than 49: $$\left({\frac{1}{\left(\pi-3\right)^2}}-50\right)^2< \left(\frac{1}{\left(\pi-3\right)^2}-7^2\right)^2$$ so $1+\frac{1}{15\sqrt{2}}$ is closer to $\frac{\pi}{3}$ than $1+\frac{1}{21}$ is.
$$\frac{\pi}{3}\approx 1.04719$$
$$1+\frac{1}{15\sqrt{2}}\approx 1.0471(4)$$
$$1+\frac{1}{21}\approx 1.047(6)$$
In other words,$$\left(\pi-3\right)^2\approx0.020048\approx0.02=\frac{1}{50}$$ so $$\pi-3\approx\frac{1}{\sqrt{50}}=\frac{1}{5\sqrt{2}}$$
Equivalently, $$\frac{\pi}{3}\approx1+\frac{1}{15\sqrt{2}}$$ An integral for $\frac{22}{7}-\pi$ is given by
$$\frac{22}{7}-\pi=\int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}dx$$
and a series for $\frac{\pi}{3}-1$ is given by
$$\frac{\pi}{3}-1=\sum_{k=1}^{\infty}\frac{1}{\left(1+k\right)\left(1+2k\right)\left(1+4k\right)}$$ Q: Is there a similar integral or series for $\frac{\pi}{3}-\left(1+\frac{1}{15\sqrt{2}}\right)$?
[EDIT 4/02/2016] Q2: Is there a variant of $$\pi=\frac{8}{3}\sum_{k=1}^{\infty}\frac{sin\left(\frac{k\pi}{4}\right)}{k}$$ that truncates to $3+\frac{1}{5\sqrt{2}}$?
For instance, there is $$\pi=\frac{1}{8944} \sum_{k=0}^\infty \left(\frac{77810+903\sqrt{2}}{8k+1}-\frac{131520}{8k+2}+\frac{9870+903\sqrt{2}}{8k+3}+\frac{43840}{8k+4}+\frac{77810-903\sqrt{2}}{8k+5}-\frac{131520}{8k+6}+\frac{9870-903\sqrt{2}}{8k+7}+\frac{43840}{8k+8}\right)$$ with first term $3+\frac{1}{5\sqrt{2}}$ but there may be a simpler one.
This was earlier a bit hastily closed. However, one aspect of the question might have an interesting connection to the Tribonacci constant $T$. First, let $w = \frac{\sqrt{2}}{T^{-1}+1}$, so,
$$j\big(\tfrac{1+\sqrt{-11}}{2}\big) = \frac{(w^{24}-16)^3}{w^{24}} = -2^{15}$$
where $j(\tau)$ is the j-function. We then get the Ramanujan/Chudnovsky-type pi formula,
$$\frac{1}{\pi} = 4\sum_{n=0}^\infty (-1)^n\frac{(6n)!}{(3n)!\,n!^3}\frac{154n+15}{(2^{15})^{n+1/2}}\tag1$$
The first term of this just so happens to be,
$$\frac{1}{\pi} \approx \frac{4\times 15}{\sqrt{2^{15}}} = \frac{15\sqrt{2}}{64}$$
hence,
$$\frac{\pi}{64} \approx \frac{1}{15\sqrt{2}}\tag2$$
We wish to find a connection between $\frac{\pi}{64}$ and $\frac{\pi}{3}$. Looking at the continued fraction (WA link) of $\frac{\pi}{64}$, the fourth convergent is $\color{brown}{\frac{3}{61}}$ and it turns out that,
$$\frac{\pi}{64}+\color{brown}{\frac{\pi}{64}\frac{61}{3}} = \frac{\pi}{3}$$
However, since $\displaystyle \color{brown}{\frac{\pi}{64}\frac{61}{3}} \approx 1$, then,
$$\frac{\pi}{64}+1 \approx \frac{\pi}{3}\tag3$$
So $(2)$ and $(3)$ "explains" the relatively close approximation,
$$\begin{aligned} \frac{\pi}{3} &= 1.04719\dots\\ 1+\frac{1}{15\sqrt{2}}&= 1.04714\dots \end{aligned}$$