26
$\begingroup$

Stated here without proof is the magnificent series $$\frac{48}{371} \sum_{k=0}^\infty \frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)} \\=\pi-\frac{333}{106},$$ which proves that $\pi>333/106$.

I can only assume that the series is proven using the integral $$\pi-\frac{333}{106}=\frac{1}{530}\int_0^1 \frac{x^5(1-x)^6(197+462x^2)}{1+x^2}dx.$$My attempts have been so far to split up the integral as $$\begin{align} 530J&=\int_0^1 \frac{x^5(1-x)^6(197+462x^2)}{1+x^2}dx\\ &=197\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx+462\int_0^1\frac{x^{7}(1-x)^6}{1+x^2}dx\\ &=197J_1+462J_2. \end{align}$$ Each remaining integral is turned into a series with $$\frac1{1+x^2}=\sum_{n\ge0}(-1)^n x^{2n}$$ so we have two series of the form $$f(p)=\sum_{n\ge0}(-1)^n\int_0^1 x^{p+2n}(1-x)^6dx=720\sum_{n\ge0}(-1)^n\frac{(p+2n)!}{(p+2n+7)!}.$$ Each factorial term is rewritten as $$\frac{s!}{(s+7)!}=\frac1{(s+1)(s+2)(s+3)(s+4)(s+5)(s+6)(s+7)},$$ so that $$f(p)=720\sum_{n\ge0}\frac{(-1)^n}{\prod_{k=1}^{7}(2n+p+k)}.$$ Then $$J_1=f(5)\\ J_2=f(7).$$ But how does one get from $530J=197f(5)+462f(7)$ to the series in question? Furthermore, how do we prove that $J=\pi-333/106$? I would assume that one would at some point use the binomial theorem then be left with a bunch of integrals like $$\int_0^1\frac{x^qdx}{1+x^2}$$ which I suppose are evaluable in terms of $\pi$, but it seems like a great deal of cancellation/simplification would have to occur and I do not immediately see where this would happen. There has to be an easier way.

Thanks!

$\endgroup$
8
  • 2
    $\begingroup$ Interesting post, for sure. I do not see how $$\int_0^1\frac{x^qdx}{1+x^2}=\frac{1}{4} \left(\psi\left(\frac{q+3}{4}\right)-\psi \left(\frac{q+1}{4}\right)\right)$$ could be are evaluated in terms of $\pi$. $\endgroup$ Commented Dec 17, 2019 at 9:13
  • 2
    $\begingroup$ @ClaudeLeibovici Use the digamma reflection formula. $\endgroup$ Commented Dec 17, 2019 at 22:14
  • 1
    $\begingroup$ @SimplyBeautifulArt. Shame on me ! I forgot this one. $\endgroup$ Commented Dec 18, 2019 at 4:10
  • $\begingroup$ @SimplyBeautifulArt how may we use the digamma reflection to simplify $$\psi_0(\tfrac{q+3}{4})-\psi_0(\tfrac{q+1}{4})?$$ this is of the form $$\psi_0(s+\tfrac12)-\psi_0(s)...$$ $\endgroup$
    – clathratus
    Commented Dec 18, 2019 at 20:39
  • $\begingroup$ @clathratus You can reduce it using the recurrence relation (i.e. polynomial long division) to $\psi(3/4)-\psi(1/4)$, to which you can apply the reflection formula (i.e. solving $\int_0^1\frac{\mathrm dx}{1+x^2}$ and $\int_0^1\frac{x~\mathrm dx}{1+x^2}$). $\endgroup$ Commented Dec 18, 2019 at 20:45

2 Answers 2

13
$\begingroup$

Probably not a very elegant method, as it makes a great use of a CAS. However it seems quite general for this kind of series. Inversely, it can be used to create similar results.

The terms of the series, which are a rational function of the index, can be decomposed into a sum of rational terms \begin{equation} u_k=\sum_{j=1}^n\frac{\lambda_j}{k+a_j} \end{equation} (we suppose the order of the poles is 1). When $\left|x\right|<1$, the series \begin{equation} f_j(x)=\sum_{k=0}^\infty \frac{x^k}{k+a_j}=x^{-a_j}\int_0^x \frac{t^{a_j}}{1-t}\,dt+\frac{1}{a_j} \end{equation} This can be verified by developping the $(1-t)^{-1}$ term in the integral. Then, the series \begin{align} S(x)&=\sum_{k=0}^\infty u_kx^k\\ &=\sum_{k=0}^\infty\sum_{j=1}^n\frac{\lambda_jx^k}{k+a_j}\\ &=\sum_{j=1}^n\lambda_j\left[x^{-a_j}\int_0^x \frac{t^{a_j}}{1-t}\,dt+\frac{1}{a_j}\right]\\ &=\sum_{j=1}^n\frac{\lambda_j}{a_j}+\int_0^x \frac{\sum_{j=1}^n\lambda_jx^{-a_j}t^{a_j}}{1-t}\,dt \end{align} The proposed series corresponds to $\lim_{x\to1^{-}}S(x)$. Due to the denominator in the integral, in order that this limit exists, the condition \begin{equation} \sum_{j=1}^n\lambda_j=0 \end{equation} must hold. Then \begin{equation} S(1)=\sum_{j=1}^n\frac{\lambda_j}{a_j}+\int_0^1 \frac{\sum_{j=1}^n\lambda_jt^{a_j}}{1-t}\,dt \end{equation} The remaining integral can be directly calculated.

In the proposed case, using a CAS, \begin{align} u_k&=\frac{48}{371}\frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)} \\ &=-{\frac {181203}{3799040\,k+21844480}}+{\frac {418643}{759808\,k+ 3229184}}-{\frac {293677}{759808\,k+2849280}}\\ &\,\quad+{\frac {743573}{3799040 \,k+12346880}}+{\frac {181203}{759808\,k+3988992}}-{\frac {1868267}{ 3799040\,k+18045440}}\\ &\,\quad-{\frac {56237}{759808\,k+2089472}}+{\frac {56237 }{3799040\,k+8547840}} \end{align} after some calculations, one obtains \begin{align} \sum_{j=1}^n\frac{\lambda_j}{a_j}&=\frac{7516928}{124151182155}\\ \sum_{j=1}^n\lambda_jt^{a_j}&=-{\frac {181203}{3799040}{t}^{{\frac{23}{4}}}}+{\frac {418643}{759808} {t}^{{\frac{17}{4}}}}-{\frac {293677}{759808}{t}^{{\frac{15}{4}}}}+{ \frac {743573}{3799040}{t}^{{\frac{13}{4}}}}\\ &\,\quad+{\frac {181203}{759808}{t }^{{\frac{21}{4}}}}-{\frac {1868267}{3799040}{t}^{{\frac{19}{4}}}}-{ \frac {56237\,{t}^{11/4}}{759808}}+{\frac {56237\,{t}^{9/4}}{3799040}}\\ &=\frac{1}{3799040} \left( 181203t+56237 \right)t^{9/4}\left( 1-\sqrt{t} \right)^5 \end{align} The above function vanishes at $t=1$, as expected. We have to evaluate \begin{align} S(1)&=\frac{7516928}{124151182155}+\frac{1}{3799040} \int_0^1 \frac{\left( 181203t+56237 \right)t^{9/4}\left( 1-\sqrt{t} \right)^5}{1-t}\,dt \\ &=\frac{7516928}{124151182155} +\frac{1}{949760}\int_0^1\frac{\left( 181203v^4+56237 \right)v^{12}\left( 1-v^2 \right)^5}{1+v^2}\,dv \end{align} To evaluate the integral, by devloping the numerator, we have to calculate terms as \begin{equation} I_n=\int_0^1\frac{v^{2n}}{1+v^2}\,dv \end{equation} A recurrence relation can be found easily: \begin{equation} I_n=\frac{1}{2n-1}-I_{n-1} \end{equation} from which we have (with $I_0=\pi/4$) \begin{equation} I_n=(-1)^{n-1}\sum_{p=0}^{n-1}\frac{(-1)^{p}}{2p+1}+(-1)^n\frac{\pi}{4} \end{equation} After (rather uninteresting) calculations, we get \begin{equation} \frac{1}{949760}\int_0^1\frac{\left( 181203v^4+56237 \right)v^{12}\left( 1-v^2 \right)^5}{1+v^2}\,dv=\pi-{\frac{780059253811}{248302364310}} \end{equation} Finally \begin{equation} S(1)=\pi-\frac{333}{106} \end{equation} as expected.

$\endgroup$
4
  • $\begingroup$ Really really nice job. Very illuminating (+1) $\endgroup$
    – clathratus
    Commented Dec 17, 2019 at 17:31
  • $\begingroup$ @clathratus Thanks for this comment! $\endgroup$
    – Paul Enta
    Commented Dec 17, 2019 at 19:28
  • 2
    $\begingroup$ This would be almost impossible to do by hand. $\endgroup$ Commented Dec 21, 2019 at 23:43
  • 1
    $\begingroup$ @martycohen I must agree. I think though, given the caliber of the series in question, purely manual calculation is an unreasonable expectation. Fell free to prove me wrong though ;) $\endgroup$
    – clathratus
    Commented Dec 22, 2019 at 5:58
8
$\begingroup$

This series can be obtained with the same technique used in A series to prove $\frac{22}{7}-\pi>0$

Let us start from series $$\sum _{k=0}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=\pi-\frac{64}{21}$$

to obtain the following truncations:

$$\sum _{k=1}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=\pi-\frac{2176}{693}$$ $$\sum _{k=2}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=\pi-\frac{4288}{1365}$$ $$\sum _{k=3}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=\pi-\frac{45708032}{14549535}$$

The approximation we are interested in lies between two of these fractions.

$$ \frac{4288}{1365}< \frac{333}{106} < \frac{45708032}{14549535}$$

Therefore, a series for $\pi-\frac{333}{106}$ can be obtained as a mix of the series for $\pi-\frac{4288}{1365}$ and $\pi-\frac{45708032}{14549535}$.

From $$\pi-\frac{333}{106} = a(\pi-\frac{4288}{1365})+b(\pi-\frac{45708032}{14549535})$$

we obtain $$a=\frac{56237}{237440}$$ $$b=\frac{181203}{237440}$$

Finally,

$$\pi-\frac{333}{106}=\frac{56237}{237440}(\pi-\frac{4288}{1365})+\frac{181203}{237440}(\pi-\frac{45708032}{14549535})=$$ $$\frac{56237}{237440}\sum _{k=2}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}+$$ $$\frac{181203}{237440}\sum _{k=3}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=$$ $$\frac{56237}{237440}\sum _{k=0}^\infty \frac{960}{(4k+9)(4k+11)(4k+13)(4k+15)(4k+17)(4k+19)}+$$ $$\frac{181203}{237440}\sum _{k=0}^\infty \frac{960}{(4k+13)(4k+15)(4k+17)(4k+19)(4k+21)(4k+23)}=$$ $$\frac{48}{371} \sum_{k=0}^\infty \frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)}$$

$\endgroup$
11
  • $\begingroup$ Very interesting. How do you know the initial series $\pi-64/21$? $\endgroup$
    – Paul Enta
    Commented Dec 21, 2019 at 23:41
  • 1
    $\begingroup$ I start from $$\frac{\pi}{8}=\sum_{k=0}^\infty \frac{1}{(4k+1)(4k+3)}$$ and plug more factors into WolframAlpha. $\endgroup$ Commented Dec 21, 2019 at 23:44
  • 1
    $\begingroup$ Thank you both! You may want to have a look at what FDP managed to get: a linear numerator. math.stackexchange.com/questions/1656475/… The remaining question is: is there a similar series with constant numerator? $\endgroup$ Commented Dec 22, 2019 at 6:23
  • 1
    $\begingroup$ Yes, but this ensures positive coefficients a and b, which leads to series with all terms positive. This proves directly that pi is larger than the fraction. $\endgroup$ Commented Dec 22, 2019 at 15:36
  • 1
    $\begingroup$ Interesting, but this is not how I get it. I try series with this kind of denominators on WolframAlpha and I write down the ones I like for some reason. Here, we have int(21pi)=64, so the series is somehow interesting. Anyway, maybe you can get this one from five pieces, with denominators (4k+1)(4k+3), (4k+3)(4k+5), ... ( $\endgroup$ Commented Dec 26, 2019 at 7:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .