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.


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 $9.9$?

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    $\begingroup$ Actually, it's not that close... $\endgroup$ – Surb Feb 8 '16 at 18:31
  • $\begingroup$ @Surb: Or "what's the distance from $\pi^2$ to $10$?" The answer would be the same. $\endgroup$ – Jaume Oliver Lafont Feb 8 '16 at 18:39
  • $\begingroup$ Why do you even mention Noam Elkies' web page, since you don't intend to use anything from it to improve your approximation? $\endgroup$ – David K Feb 8 '16 at 19:05
  • $\begingroup$ @DavidK It is the same problem, it is the paper that motivated me to try a different solution and there are also telescoping series in the proof of the series I wrote without a proof. Maybe I should add that proof. $\endgroup$ – Jaume Oliver Lafont Feb 8 '16 at 21:48
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    $\begingroup$ I didn't realize you were citing that paper as inspiration rather than as a possible source of facts to use. As a source of inspiration to try your own approach, that's fine. $\endgroup$ – David K Feb 8 '16 at 22:06

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.

$$\begin{align} \pi^2 &= 9 + \sum_{k=0}^\infty \frac{3}{(k + 1)^2 (k + 2)^2}\\ &=\frac{39}{4} + \sum_{k=1}^\infty \frac{3}{(k + 1)^2 (k + 2)^2}\\ \\ \pi^2 &= \frac{21}{2} - \sum_{k=0}^\infty \frac{6}{(k + 1) (k + 2)^2 (k+3)}\\ &=10 -\sum_{k=1}^\infty \frac{6}{(k + 1) (k + 2)^2 (k+3)}\\ \end{align}$$

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    $\begingroup$ Why are you asking and answering your own question immediately? $\endgroup$ – BigbearZzz Feb 7 '16 at 6:55
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    $\begingroup$ Maybe he forgot to sign out before answering on another account. $\endgroup$ – Decaf-Math Feb 7 '16 at 6:56
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    $\begingroup$ Maybe he want to earn the "Self-Learner" badge? xd $\endgroup$ – Mythomorphic Feb 7 '16 at 7:15
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    $\begingroup$ @BigbearZzz Related (and showing your puzzlement is legitimate). $\endgroup$ – Did Feb 7 '16 at 7:54
  • $\begingroup$ @BigbearZzz I had an answer for a question I did not find, so I added both myself. $\endgroup$ – Jaume Oliver Lafont Feb 8 '16 at 22:23

There is really no incredibly complex reason for this. It is called a mathematical coincidence. Your question is like asking:

"Why is $e^\pi -\pi\approx20$?"


An integral proof that $10>\pi^2$ is given by

$$\int_0^1 \frac{24 x^2 (1 - x)^2 (49 - 100 x (1 - x)^2) \log(\frac{1}{x})}{151 (1 + x^2)} dx = 10 - \pi^2$$

WolframAlpha link


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