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What would be an easy method to find the approximate value of/close the form (the former will work too, if reasonably correct, a few decimal places, not more than that.): $$\sum_{n=1}^{\infty}\frac{(2n+99)!(3n-2)!}{(2n)!(3n+99)!}$$ I thought we could write it as: $$\frac{99!}{101!}\sum_{n=1}^{\infty}\frac{(2n+99)!(3n-2)!(101)!}{(2n)!(99)!(3n+99)!}=\frac{1}{101\times100}\sum_{n=1}^{\infty}\frac{\displaystyle\binom{2n+99}{99}}{\displaystyle\binom{3n+99}{101}}$$ I don't know of a simple way to proceed.

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  • $\begingroup$ Would this at least partially answer your question ? $\endgroup$ – Lucian May 10 '15 at 13:38
  • $\begingroup$ Mathematica's Sum[] function gives $5/1000$ approximately. $\endgroup$ – JamalS May 10 '15 at 13:43
  • $\begingroup$ @Lucian So that's the exact result? $\endgroup$ – Kitegi May 10 '15 at 14:27
  • $\begingroup$ @Farnight: Yes. $\endgroup$ – Lucian May 10 '15 at 14:36
  • $\begingroup$ where is this coming from!? $\endgroup$ – Math-fun May 11 '15 at 10:34
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For any $s \ge 1$, let $$\Delta_s = \sum_{n=1}^\infty \frac{(2n+s-1)!}{(2n)!}\frac{(3n-2)!}{(3n+s-1)!}$$ In particular, $\Delta_{100}$ is the sum we want to compute. Notice

$$ \begin{align} \frac{(2n+s-1)!}{(2n)!} &= \left.\left(\frac{d}{dz}\right)^{s-1} z^{2n+s-1}\right|_{z=1} = \frac{\Gamma(s)}{2\pi i}\oint_{C} \frac{z^{2n+s-1}}{(z-1)^s}dz\\ \frac{(3n-2)!}{(3n+s-1)!} &= \frac{1}{\Gamma(s+1)}\frac{\Gamma(3n-1)\Gamma(s+1)}{\Gamma(3n+s)} = \frac{1}{\Gamma(s+1)}\int_0^1 t^{3n-2} (1-t)^s dt \end{align}$$ where $C_1 \subset \mathbb{C}$ is a small circular contour centered at $z = 1$. We have following integral representation for $\Delta_s$.

$$s\Delta_s = \frac{1}{2\pi i}\int_0^1 \oint_C \left(\frac{z(1-t)}{z-1}\right)^s \frac{zt}{1 - z^2t^3} dz dt $$

Let $\Delta(\eta)$ be the OGF for $s\Delta_s$, $$\Delta(\eta) \stackrel{def}{=} \sum_{s=1}^\infty s\Delta_s \eta^{s-1}$$ Let $\omega = (1-\eta + \eta t)^{-1}$ and $C_{\omega} \subset \mathbb{C}$ be a circular contour centered at $\omega$. Since

$$\sum_{s=1}^\infty \left(\frac{z(1-t)}{z-1}\right)^s \eta^{s-1} = \frac{z(1-t)}{z(1-\eta + \eta t) - 1} = \frac{z(1-t)\omega}{z-\omega} $$ We find $$\begin{align} \Delta(\eta) &= \frac{1}{2\pi i}\int_0^1 \oint_{C_\omega} \frac{z^2 t (1-t)}{1 - z^2 t^3}\frac{\omega}{z - \omega} dz dt = \int_0^1 \frac{\omega^3 t(1-t)}{1 - \omega^2 t^3} dt\\ &= \int_0^1 \frac{ t dt }{(1-\eta +\eta t)((1-\eta)^2 + (1-\eta^2) t + t^2)}\\ &= \frac{1}{1-\eta}\int_0^{\frac{1}{1-\eta}} \frac{t dt}{(1+\eta t)(1 + (1+\eta)t + t^2)}\\ &= \frac{1}{(1-\eta)^2}\int_0^{\frac{1}{1-\eta}} \left[\frac{t+\eta}{1 + (1+\eta)t + t^2} - \frac{\eta}{1+\eta t}\right] dt\\ &= \frac{1}{(1-\eta)^2}\left[ \frac12 \log(3-2\eta) - \sqrt{\frac{1-\eta}{3+\eta}}\tan^{-1}\left(\frac{\sqrt{(1-\eta)(3+\eta)}}{3-\eta}\right) \right] \end{align} $$ Throw the last expression to a CAS and ask it to compute the coefficient of $\eta^{s-1}$ for $s = 100$. We get

$$ \Delta_{100} = \frac{1}{100}\left( 50 \log(3) - A \frac{\pi}{3^{199/2}} - \frac{B}{C}\right)$$ where $$\begin{array}{rcl} A &=& 279620009275010140018538432376916760970649320550\\ B &=& 402992750960761749592608273159927666000520075786\\ & & 76797445589033694041389862879632396903\\ C &=& 782839087808164519964649481706120884210784218785\\ & & 090048580229772278114461980652430352 \end{array} $$ If one multiply this expression by $101 \times 100$, one reproduce the number in Lucian's comment (which is off by above factor).

Numerically, $\Delta_{100}$ is very close to $\frac{1}{200}$,

$$\Delta_{100} \approx 0.004999999999999999999533961997106313117\ldots$$

There is a good reason for this. For small $\delta$, we have

$$ \Delta(1-\delta) = \frac{1}{\delta^2}\left(\frac12\log(1+2\delta) - \sqrt{\frac{\delta}{4-\delta}}\tan^{-1}\left(\frac{\sqrt{\delta(4-\delta)}}{2+\delta}\right)\right) \approx \frac{1}{2\delta} + O(1) $$ This implies $\Delta(\eta)$ has a simple pole at $\eta = 1$ with residue $-\frac12$. Let us subtract this pole from $\Delta(\eta)$ and look for the singularities for the remaining pieces nearest the origin.

  • For the piece $\log(3 - 2\eta)$, the nearest singularity is clearly $\eta = \frac32$.

  • For the piece $\displaystyle\;\sqrt{\frac{1-\eta}{3+\eta}}\tan^{-1}\left(\frac{\sqrt{(1-\eta)(3+\eta)}}{3-\eta}\right)\;$ the nearest singularity occurs at those $\eta$ where $\displaystyle\;\frac{\sqrt{(1-\eta)(3+\eta)}}{3-\eta} = \pm i\;$. Once again, this leads to $\eta = \frac32$.

What this means is after we subtract the pole from $\Delta(\eta)$, the remaining piece is analytic over the circle $|z| < \frac32$. If we pick a circle with radius $1 < r < \frac32$ and extract the power series of the remaining piece at $z = 0$ using Cauchy integral formula, the coefficients of $\eta^{s-1}$ for the remaining piece will grow at most as fast as $r^{1-s}$.

Using this, we can conclude for large $s$,

$$s \Delta_s = \frac12 + O( r^{1-s} ), \quad 1 < r < \frac32.$$

In order to estimate the coefficient in front of the $O(r^{1-s})$ term, we need to study $\Delta(\eta)$ near $\eta = \frac32$. One can show that the leading singular behavior there is given by

$$\Delta(\eta) \approx \frac{4}{3} \log\left(\frac32 - \eta\right) + \cdots \quad\text{ for }\quad \eta \approx \frac32.$$

This suggests for large $s$,

$$s\Delta_s \approx \frac12 - \frac{4}{3(s-1)} (2/3)^{s-1}$$

Let use $s = 100$ as a test case, this estimate gives

$$\left[\frac{100 \Delta_{100} - \frac12}{(2/3)^{99}}\right]_{\verb/approx/} = -\frac{4}{297} \approx -0.01346801346801347$$

This is within $7\%$ from corresponding exact value. Our hand waving estimate turns out to be reasonably decent (at least for $s \approx 100$).

$$\left[ \frac{100 \Delta_{100} - \frac12}{(2/3)^{99}}\right]_{\verb/exact/} \approx -0.012631530615502$$

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  • $\begingroup$ Very nice answer! +1 $\endgroup$ – Markus Scheuer May 11 '15 at 10:04
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$$\begin{align} S_p&=\sum_{n=1}^p \frac{(2n+99)\dots(2n+1)}{(3n+99)\dots(3n-1)}\\ S_p&=\sum_{n=1}^p \sum_{k=-1}^{99}\frac{a_k}{3n+k} \end{align}$$ Where: $$a_k=\frac{[2\left(\frac{-k}{3}\right)+99]\dots[2\left(\frac{-k}{3}\right)+1]}{[3\left(\frac{-k}{3}\right)+99]\dots[3\left(\frac{-k}{3}\right)+k-1][3\left(\frac{-k}{3}\right)+k+1]\dots[3\left(\frac{-k}{3}\right)-1]}$$

All of the $a_k$ can be calculated precisely since they're just rational numbers. But I don't have a way of doing it.

If we can find an approximation of $\sum_{n=1}^p \frac{1}{3n+k}$ to the order of $o(1)$, we can calculate the result.

To do that, we only need an approximation of $\sum_{n=1}^p \frac{1}{3n}$, $\sum_{n=1}^p \frac{1}{3n+1}$, and $\sum_{n=1}^p \frac{1}{3n+2}$

$$\sum_{n=1}^p \frac{1}{3n}=\frac\gamma3 + \frac{\ln(p)}{3}+o(1)$$

$$\sum_{n=1}^p \frac{1}{3n+1}=\frac{-\psi(4/3)}{3}+\frac{\ln(p)}{3}+o(1)$$

$$\sum_{n=1}^p \frac{1}{3n+2}=\frac{-\psi(5/3)}{3}+\frac{\ln(p)}{3}+o(1)$$

Where $$\psi(4/3)=3-\gamma-\frac{\pi}{2\sqrt3}-\frac{3 \ln(3)}2$$ $$\psi(5/3)=\frac32-\gamma+\frac{\pi}{2\sqrt3}-\frac{3 \ln(3)}2$$

Now we have

$$\begin{align} S_p&=\sum_{n=1}^{p}\frac{a_0+a_3+\dots+a_{99}}{3n}+\frac{a_1+a_4+\dots+a_{97}}{3n+1}+\frac{a_{-1}+a_2+\dots+a_{98}}{3n+2}\\ &-\left(a_3\frac{1}{3}+a_6\left(\frac{1}{3}+\frac{1}{6}\right)+\dots+a_{99}\left(\frac{1}{3}+\frac{1}{6}+\dots+\frac{1}{99}\right)\right)\\ &-\left(a_4\frac{1}{4}+a_7\left(\frac{1}{4}+\frac{1}{7}\right)+\dots+a_{97}\left(\frac{1}{4}+\frac{1}{7}+\dots+\frac{1}{97}\right)\right)\\ &-\left(a_5\frac{1}{5}+a_8\left(\frac{1}{5}+\frac{1}{8}\right)+\dots+a_{98}\left(\frac{1}{5}+\frac{1}{8}+\dots+\frac{1}{98}\right)\right)\\ &+a_{-1}\frac{1}{2}+o(1)\\ S_p&=\frac{(a_0+a_3+\dots+a_{99})\gamma}{3}\\ &+\frac{-(a_1+a_4+\dots+a_{97})\psi(4/3)}{3}\\ &+\frac{-(a_{-1}+a_2+\dots+a_{98})\psi(5/3)}{3}\\ &-\left(a_3\frac{1}{3}+a_6\left(\frac{1}{3}+\frac{1}{6}\right)+\dots+a_{99}\left(\frac{1}{3}+\frac{1}{6}+\dots+\frac{1}{99}\right)\right)\\ &-\left(a_4\frac{1}{4}+a_7\left(\frac{1}{4}+\frac{1}{7}\right)+\dots+a_{97}\left(\frac{1}{4}+\frac{1}{7}+\dots+\frac{1}{97}\right)\right)\\ &-\left(a_5\frac{1}{5}+a_8\left(\frac{1}{5}+\frac{1}{8}\right)+\dots+a_{98}\left(\frac{1}{5}+\frac{1}{8}+\dots+\frac{1}{98}\right)\right)\\ &+a_{-1}\frac{1}{2}+o(1)\\ \end{align}$$

This is as far as I could go. There could be some relations between the $a_k$ that'll make this simpler but I can't think of anything. (although, according to wolfram alpha, the sum seems to be close to $1/200$)

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This is the answer given by Lucian:

$$\sum_{n=1}^{\infty}\frac{(2n+99)!(3n-2)!}{(2n)!(3n+99)!}=\frac{1}{204} {_6F_5}(\{\frac23,1,1,\frac43,51,\frac{103}{2}\},\{\frac32,2,\frac{103}{3},\frac{104}{3},35\},1)$$

an evaluation of which would be

$$0.0049999999999999999995339619971063131170085423478892$$

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