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I have the following double summations:

Sum 1 : $\sum _{p=0}^{k-1} \left(\frac{\sqrt{\frac{(p+1) \Gamma \left(p+\frac{11}{4}\right)}{\Gamma (p+2)}}}{(p+2) \sqrt{\Gamma \left(\frac{11}{4}\right)}}-\sum _{j=p+2}^k \frac{\sqrt{\frac{\Gamma \left(j+\frac{7}{4}\right)}{\Gamma (j+1)}}}{\sqrt{j} (j+1) \sqrt{\Gamma \left(\frac{11}{4}\right)}}\right)$

Sum 2: $\sum _{p=0}^{k-1} \left(\frac{\sqrt{\frac{(p+1) \Gamma \left(p+\frac{11}{4}\right)}{\Gamma (p+2)}}}{(p+2) \sqrt{\Gamma \left(\frac{11}{4}\right)}}-\sum _{j=p+2}^k \frac{\sqrt{\frac{\Gamma \left(j+\frac{7}{4}\right)}{\Gamma (j+1)}}}{\sqrt{j} (j+1) \sqrt{\Gamma \left(\frac{11}{4}\right)}}\right)^2$

I empirically (and analytically) verified that Sum 1 converges as $k \to \infty$ but it wasn't clear whether Sum 2 converges.

I want to check the same using Mathematica. I tried using SumConvergence but it displayed the same output as input.

Any thoughts are appreciated.

Edit - 1: Mathematica Input Forms:

Sum[(Sqrt[(p + 1)*(Gamma[11/4 + p]/(Gamma[11/4] * Gamma[2 + p]))]/(p + 2)) - Sum[ Sqrt[Gamma[7/4 + j]/(Gamma[11/4]*Gamma[1 +j])]/(Sqrt[j]*(j + 1)), {j,p + 2, k}] , {p, 0, k - 1}]

Sum[((Sqrt[(p + 1)*(Gamma[11/4 + p]/(Gamma[11/4] * Gamma[2 + p]))]/(p + 2)) - Sum[ Sqrt[Gamma[7/4 + j]/(Gamma[11/4]*Gamma[1 +j])]/(Sqrt[j]*(j + 1)), {j,p + 2, k}])^2 , {p, 0, k - 1}]

Edit 2: Some background about this expression based on @user64494 answer:


I have a random walk with the following expression

$S = \frac{1}{2} x_0+\frac{\sqrt{2}}{3} \sqrt{1+\frac{r}{2}} \left(x_1-\frac{x_0}{2}\right)+\frac{\sqrt{3}}{4} \sqrt{(1+\frac{r}{2})(1+\frac{r}{3})} \left(x_2-\frac{x_0+x_1}{3}\right)+\frac{\sqrt{4}}{5} \sqrt{(1+\frac{r}{2})(1+\frac{r}{3})(1+\frac{r}{4})} \left(x_3-\frac{x_0+x_1+x_2}{4}\right) + \dots$

with $0<r<1$ and $x_0,x_1,x_2, \dots $ are sampled independently from Gaussian distribution with mean $1$ and variance $1$. I want to show that $\Bbb{E}(S)$ and $Var(S)$ are finite.


To see that Sum 1 converges, consider the following:

Let $r = 0.75$ and let $x_0 = x_1 = x_2 \dots = 1$.

Then $S = \sum_{p=0}^{\infty} \frac{\sqrt{\Pi_{j=2}^{(p+1)} (1+\frac{0.75}{j}) }}{\sqrt{(p+1)}{(p+2)}} $

Now, Product[1 + (3/(4*j)), {j, 2, p + 1}] is $\frac{\Gamma \left(p+\frac{11}{4}\right)}{\Gamma \left(\frac{11}{4}\right) \Gamma (p+2)}$ which, after taking the square-root, is $O(p^{3/8})$ as $p \to \infty$. Therefore, each term in the above series is $O(p^{-9/8})$ and hence the series converges.

Now Sum 1 follows from rearranging $S$ by collecting each $x_i$ separately and setting $r = 0.75$ and let $x_0 = x_1 = x_2 \dots = 1$.

This can be seen as follows -

$S = \frac{1}{2} x_0+\frac{\sqrt{2}}{3} \sqrt{1+\frac{r}{2}} \left(x_1-\frac{x_0}{2}\right)+\frac{\sqrt{3}}{4} \sqrt{(1+\frac{r}{2})(1+\frac{r}{3})} \left(x_2-\frac{x_0+x_1}{3}\right)+\frac{\sqrt{4}}{5} \sqrt{(1+\frac{r}{2})(1+\frac{r}{3})(1+\frac{r}{4})} \left(x_3-\frac{x_0+x_1+x_2}{4}\right) + \dots + \frac{\sqrt{k}}{k+1} \sqrt{(1+\frac{r}{2})(1+\frac{r}{3})(1+\frac{r}{4})\dots(1+\frac{r}{k})} \left(x_{k-1}-\frac{x_0+x_1+x_2+\dots+x_{k-2}}{k}\right)$

By collecting the $x_i$'s separately, we have

$S = x_0\left(\frac{1}{2} - \left(\frac{\sqrt{1+\frac{r}{2}}}{\sqrt{2}\times 3} +\frac{\sqrt{(1+\frac{r}{2})(1+\frac{r}{3})}}{\sqrt{3}\times 4} + \frac{\sqrt{(1+\frac{r}{2})(1+\frac{r}{3})(1+\frac{r}{4})}}{\sqrt{4}\times 5} + \dots \right) \right) + x_1\left(\frac{\sqrt{2}}{3} \sqrt{1+\frac{r}{2}} - \left(\frac{\sqrt{(1+\frac{r}{2})(1+\frac{r}{3})}}{\sqrt{3}\times 4} + \frac{\sqrt{(1+\frac{r}{2})(1+\frac{r}{3})(1+\frac{r}{4})}}{\sqrt{4}\times 5} + \dots \right) \right) + x_2\left(\frac{\sqrt{3}}{4} \sqrt{(1+\frac{r}{2})(1+\frac{r}{3})} - \left( \frac{\sqrt{(1+\frac{r}{2})(1+\frac{r}{3})(1+\frac{r}{4})}}{\sqrt{4}\times 5} + \dots \right) \right) + \dots + x_{k-1}\left(\frac{\sqrt{k}}{k+1} \sqrt{(1+\frac{r}{2})(1+\frac{r}{3})\dots(1+\frac{r}{k})} \right)$

i.e. $S = \sum_{p=0}^{k-1} x_p\left(\frac{\sqrt{p+1}}{p+2}\sqrt{\Pi_{i=2}^{p+1}(1+\frac{r}{i}) } - \left(\sum_{j=p+2}^{k} \frac{\sqrt{\Pi_{i=2}^{j}(1+\frac{r}{i}) }}{\sqrt{j}(j+1)} \right) \right) $

Now, substitute $r=3/4$ and $x_0 = x_1 = \dots = 1$; compute the two products - a) Product[1 + (3/(4*i)), {i, 2, p + 1}] = $\frac{\Gamma \left(p+\frac{11}{4}\right)}{\Gamma \left(\frac{11}{4}\right) \Gamma (p+2)}$ and b) Product[1 + (3/(4 i)), {i, 2, j}] = $\frac{\Gamma \left(j+\frac{7}{4}\right)}{\Gamma \left(\frac{11}{4}\right) \Gamma (j+1)}$.

Hence we have $ S = \sum_{p=0}^{k-1} \left(\frac{\sqrt{p+1}}{p+2}\sqrt{ \frac{\Gamma \left(p+\frac{11}{4}\right)}{\Gamma \left(\frac{11}{4}\right) \Gamma (p+2)} } - \left(\sum_{j=p+2}^{k} \frac{\sqrt{\frac{\Gamma \left(j+\frac{7}{4}\right)}{\Gamma \left(\frac{11}{4}\right) \Gamma (j+1)}}}{\sqrt{j}(j+1)} \right) \right)$

The reason I am rearranging in the form of Sum 1 is because I want to check for the convergence of Variance, which is closely related to Sum 2, which in turn is related to Sum 1.

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    $\begingroup$ Please edit your question to include the Mathematica code (InputForm) for these sums so that it can be copied and pasted into a notebook. $\endgroup$
    – Bob Hanlon
    Oct 1 '21 at 14:53
  • $\begingroup$ @BobHanlon : Thank you for pointing it. I added them. $\endgroup$ Oct 1 '21 at 15:05
  • $\begingroup$ Series[Sqrt[ Gamma[7/4 + j]/(Gamma[11/4]*Gamma[1 + j])]/(Sqrt[j]*(j + 1)), {j, Infinity, 2}] results in $$O\left(\left(\frac{1}{j}\right)^{17/8}\right)+\frac{\left(\frac{1}{j}\right)^{9/8}}{\sqrt{\Gamma \left(\frac{11}{4}\right)}}.$$ $\endgroup$
    – user64494
    Oct 1 '21 at 16:55
  • $\begingroup$ Can you explain the relation beween $S = \sum_{p=0}^{\infty} \frac{\sqrt{\Pi_{j=2}^{(p+1)} (1+\frac{0.75}{j}) }}{\sqrt{(p+1)}{(p+2)}}$ in your Edit 2 and Sum 1, giving us details? TIA. $\endgroup$
    – user64494
    Oct 1 '21 at 18:38
  • $\begingroup$ @user64494: Thank you for your comments. May I request to kindly not delete your existing answer in future because it brings out well the crux of this problem. $\endgroup$ Oct 1 '21 at 18:57
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Despite your claim, Sum 1 diverges. Here are my arguments. First,

Series[Sqrt[Gamma[7/4 + j]/(Gamma[11/4]*Gamma[1 + j])]/(Sqrt[j]*(j+ 1)), {j,Infinity,2}]

$$\frac{\left(\frac{1}{j}\right)^{9/8}}{\sqrt{\Gamma \left(\frac{11}{4}\right)}}+O\left(\left(\frac{1}{j}\right)^{17/8}\right) $$

Second,

AsymptoticSum[(1/j)^(9/8)/Sqrt[Gamma[11/4]],{j,p+2,k},k -> Infinity, Assumptions-> p>0]

HurwitzZeta[9/8, 2 + p]/Sqrt[Gamma[11/4]]

Third,

Series[(Sqrt[(p + 1)*(Gamma[11/4 + p]/(Gamma[11/4]*Gamma[2 + p]))]/(p + 2)) - 
HurwitzZeta[9/8, 2 + p]/Sqrt[Gamma[11/4]], p -> Infinity]

$$-\frac{7 \sqrt[8]{\frac{1}{p}}}{\sqrt{\Gamma \left(\frac{11}{4}\right)}}+O\left(\left(\frac{1}{p}\right)^{9/8}\right) $$

If I am not mistaken, the latest result implies the divergence of Sum 1 as $k\to \infty$.

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  • $\begingroup$ Calculations confirm the divegence: for k=100 Sum 1 equals 2.7, for k=200 - 3.00, for k=300 - 3.16. $\endgroup$
    – user64494
    Oct 1 '21 at 17:43
  • $\begingroup$ For k=400 Sum 1 equals 3.22. $\endgroup$
    – user64494
    Oct 1 '21 at 17:49
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    $\begingroup$ I don't know if one can simply 'confirm' divergence or convergence of a series by simply summing it upto large values and then eyeballing the values. Please cite any rigorous sources for this technique, I am curious. Thanks. $\endgroup$ Oct 1 '21 at 20:15
  • $\begingroup$ @honeybadge: You are right, but the series Sum[p^(-9/8),{p,1,k}] converges with the rate k^(-1/8). Unfortunately Table[N[(1/k)^(1/8) /. k -> n], {n, 100, 400, 100}] results in {0.562341, 0.515669, 0.490185, 0.472871} showing a faster rate, than the direct calculations do. $\endgroup$
    – user64494
    Oct 1 '21 at 20:44
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This is my attempt to prove that Sum 2 also converges.

Let Sum 1 i.e. $\sum _{p=0}^{k-1} \left(\frac{\sqrt{\frac{(p+1) \Gamma \left(p+\frac{11}{4}\right)}{\Gamma (p+2)}}}{(p+2) \sqrt{\Gamma \left(\frac{11}{4}\right)}}-\sum _{j=p+2}^k \frac{\sqrt{\frac{\Gamma \left(j+\frac{7}{4}\right)}{\Gamma (j+1)}}}{\sqrt{j} (j+1) \sqrt{\Gamma \left(\frac{11}{4}\right)}}\right)$ be written as $\sum _{p=0}^{k-1} \left(A_p - \sum _{j=p+2}^k B_j \right)$

\begin{align} \sum _{p=0}^{k-1} \left(A_p - \sum _{j=p+2}^k B_j \right) &= \sum _{p=0}^{k-1} A_p - \sum _{p=0}^{k-1} \sum _{j=p+2}^k B_j \\ &= \sum _{p=0}^{k-1} A_p - \sum _{p=0}^{k-1} \sum _{j=0}^k \Bbb{1}_{(p+2\leq j)}B_j \\ &= \sum _{p=0}^{k-1} A_p - \sum _{j=0}^k \sum _{p=0}^{k-1} \Bbb{1}_{(p+2\leq j)}B_j \qquad \text{$\Bbb{1}$ is the indicator function}\\ &= \sum _{p=0}^{k-1} A_p - \sum _{j=0}^k B_j (j-1)_{+} \qquad \text{$(.)_{+}$ means only non -ve values}\\ &= \sum _{p=0}^{k-1} \left( A_p - pB_{p+1} \right) \qquad \text{Substituting $j-1 =p$ } \tag{1}\label{1}\\ &= \sum _{p=0}^{k-1} \left( \frac{\sqrt{\frac{(p+1) \Gamma \left(p+\frac{11}{4}\right)}{\Gamma (p+2)}}}{(p+2) \sqrt{\Gamma \left(\frac{11}{4}\right)}} - p\frac{\sqrt{\frac{ \Gamma \left(p+\frac{11}{4}\right)}{\Gamma (p+2)}}}{(p+2)\sqrt{(p+1)} \sqrt{\Gamma \left(\frac{11}{4}\right)}} \right) \\ &= \sum _{p=0}^{k-1} \frac{\sqrt{\frac{ \Gamma \left(p+\frac{11}{4}\right)}{\Gamma (p+2)}}}{(p+2) \sqrt{\Gamma \left(\frac{11}{4}\right)}} \left( \sqrt{p+1} - \frac{p}{\sqrt{(p+1)} } \right) \\ &= \sum _{p=0}^{k-1} \frac{\sqrt{\frac{ \Gamma \left(p+\frac{11}{4}\right)}{\Gamma (p+2)}}}{(p+2) \sqrt{(p+1)\Gamma \left(\frac{11}{4}\right)}} \\ &= \sum _{p=0}^{k-1} O(p^{-9/8}) \tag{2}\label{2}\\ \end{align}

From \ref{1} and \ref{2}, we can say that $A_p - \sum _{j=p+2}^k B_j = O(p^{-9/8})$

Now Sum2 is equal to $\sum _{p=0}^{k-1} \left(A_p - \sum _{j=p+2}^k B_j \right)^2 = \sum _{p=0}^{k-1} O(p^{-18/8})$, which converges.

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