Why is $\pi^2$ so close to $10$? Noam Elkies explained why $\pi^2=9.8696...$ is so close to $10$ using an inequality on Euler's solution to Basel problem 
$$\frac{\pi^2}{6}=\sum_{k=0}^{\infty} \frac{1}{\left(k+1\right)^2}$$
to form a telescoping series. As a better approximation, taking one more term of this series, $$\pi^2\approx 9.9$$ was obtained.
http://www.math.harvard.edu/~elkies/Misc/pi10.pdf
There is a simple series for $\pi$ whose first term is $3$, so it may be regarded as an explanation why $\pi$ is close to $3$ and a proof that $\pi>3$, because all the remaining terms are positive as well.
$$
\begin{align}
\pi
&=
3\sum_{k=0}^{\infty}\frac{1}{(4k+1)(2k+1)(k+1)} \\
&=
3+3\sum_{k=1}^{\infty}\frac{1}{(4k+1)(2k+1)(k+1)}
\end{align}
$$
Q: Is there a similar way to show that $\pi^2$ is close to $10$ and have an approximation closer than $9.9$? 

Truncating the following series by Ramanujan that converges to $\pi^2$ from above,
$$\pi^2=10-\sum_{k=0}^\infty\frac{1}{((k+1)(k+2))^3}=10-\frac{1}{8}-\frac{1}{216}-\frac{1}{1728}-\frac{1}{8000}-\frac{1}{27000}-...$$
yields
$$\pi^2<10-\frac{1}{8}=\frac{79}{8}=9.875$$
Both $10$ and $\frac{79}{8}$ are convergents from the continued fraction for $\pi^2$.
From below,
$$\pi^2=\frac{39}{4}+\sum_{k=0}^\infty\frac{4}{((k+1)(k+2)(k+3))^2}=\frac{39}{4}+\frac{1}{9}+\frac{1}{144}+\frac{1}{900}+\frac{1}{3600}+\frac{1}{11025}+...$$
taking the first term only gives
$$\pi^2>\frac{39}{4}+\frac{1}{9}=\frac{355}{36}=9.861...$$
Finally,
$$10-\frac{1}{4}<\frac{355}{36}<\pi^2<\frac{79}{8}<10$$
[EDIT]
These series have a sixth order polynomial in the denominator. The same approximations $\dfrac{39}{4}$ and $10$ are related to simpler series using fourth order polynomials.
$$\begin{align}
\pi^2 &= 9 + \sum_{k=0}^\infty  \frac{3}{(k + 1)^2 (k + 2)^2}\\
&=\frac{39}{4} + \sum_{k=1}^\infty \frac{3}{(k + 1)^2 (k + 2)^2}\\
\\
\pi^2 &= \frac{21}{2} - \sum_{k=0}^\infty  \frac{6}{(k + 1) (k + 2)^2 (k+3)}\\
&=10 -\sum_{k=1}^\infty  \frac{6}{(k + 1) (k + 2)^2 (k+3)}\\
\end{align}$$
