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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 $9.9$?

Jaume Oliver Lafont
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    Actually, it's not that close... – Surb Feb 08 '16 at 18:31
  • @Surb: Or "what's the distance from $\pi^2$ to $10$?" The answer would be the same. – Jaume Oliver Lafont Feb 08 '16 at 18:39
  • Why do you even mention Noam Elkies' web page, since you don't intend to use anything from it to improve your approximation? – David K Feb 08 '16 at 19:05
  • @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. – Jaume Oliver Lafont Feb 08 '16 at 21:48
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    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. – David K Feb 08 '16 at 22:06

3 Answers3

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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.

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

Jaume Oliver Lafont
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    Why are you asking and answering your own question immediately? – BigbearZzz Feb 07 '16 at 06:55
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    Maybe he forgot to sign out before answering on another account. – Decaf-Math Feb 07 '16 at 06:56
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    Maybe he want to earn the "Self-Learner" badge? xd – Mythomorphic Feb 07 '16 at 07:15
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    @BigbearZzz Related (and showing your puzzlement is legitimate). – Did Feb 07 '16 at 07:54
  • @BigbearZzz I had an answer for a question I did not find, so I added both myself. – Jaume Oliver Lafont Feb 08 '16 at 22:23
  • @JaumeOliverLafont Fair enough I guess? Anyway, if you already have the answer to your question why did you ask it? This somehow doesn't make much sense for me... – BigbearZzz Feb 09 '16 at 01:01
  • @BigbearZzz As the link provided by Did shows, I had done that the day before, and it was soon linked by a similar question that can be answered the same way. That makes sense, doesn't it? – Jaume Oliver Lafont Feb 09 '16 at 06:04
  • I still don't fully understand what you mean, but anyway, I guess you have your own reason and I'll respect that. – BigbearZzz Feb 09 '16 at 07:30
  • @BigbearZzz Imagine someone asks the question: "Are there series for $\pi^k- round(\pi^k)$?" This thread is $k=2$, case $k=1$ is written in http://math.stackexchange.com/a/729913/134791 and case $k=3$ or at least $k=6$ is possibly done or almost done in http://math.stackexchange.com/questions/1610024/proving-pi3-gt-31 – Jaume Oliver Lafont Feb 09 '16 at 09:01
  • @BigbearZzz Here is a related question: http://math.stackexchange.com/questions/1650436/is-pik-any-closer-to-its-nearest-integer-than-expected – Jaume Oliver Lafont Feb 15 '16 at 16:21
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There is really no incredibly complex reason for this. It is called a mathematical coincidence. Your question is like asking:

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

  • Are the series below incredibly complex? I think it would also be nice having integrals like the one for $\frac{22}{7}\approx\pi$ http://math.stackexchange.com/questions/1456033/why-do-we-need-an-integral-to-prove-that-frac227-pi

    Thank you for your answer!

     – Jaume Oliver Lafont Feb 08 '16 at 11:53
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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

Jaume Oliver Lafont
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