Sum of hypergeometric series, but I don't understand hypergeometric series.

Question

Today, a friend and I have been trying to find a general formula for the partial sums of a series that goes like this: $$ 1, 2, 8, 64, 1024, \cdots $$ we came up with a recursive formula for it: $$a(n) = 2^n\\ b(0) = 1\\ b(n>0) = a(n)\cdot b(n-1)$$ and I've managed to determine that it's some form of hypergeometric series. I've tried deciphering the Wikipedia page on the generalized hypergeometric series but there's far too much information overflow for me to properly understand it, and therefore I've been unable to find the formula we've been looking for.

What I'm asking is for

a more simplified (doesn't need to be in layman's terms, but still understandable to someone with only high-school and olympiad math experience) explanation of generalized hypergeometric series

and

what the formula is for the partial sums of the sequence, and how you got the formula.

as always, any assistance would be appreciated. Thanks.

Answer 1

You are looking for $$S_p=\sum_{n=1}^p 2^{\frac{1}{2} n(n-1) }$$ which generates the sequence $$\{1,3,11,75,1099,33867,2131019,270566475,68990043211,35253362132043\}$$ which is sequence $A181388$ in $OEIS$.

There is almost no information about it but the terms are increasing so fast$(\frac {a_{n+1}}{a_n}=2^n)$ that $$S_p\sim 2^{\frac{1}{2} p(p-1) }$$ could be more than sufficient. It would give the sequence $$\{1,2,8,64,1024,32768,2097152,268435456,68719476736,35184372088832\}$$

Answer 2

It's easy to notice that we're talking about powers of two. But which powers? The base-two logarithms of the series is $$ 0, 1, 3, 6, 10, \ldots $$ which can be identified to be the sum of the integers less than $n$. This is equal to $$ \frac{n(n-1)}{2}, \qquad n=0, 1, 2, \ldots $$ and therefore I would set the general formula to be $$ a_n = 2^{n(n-1)/2}, \qquad n = 0, 1, 2,\ldots $$

Answer 3

Just a comment on the whole thing, Let $T_n=n(n+1)/2$, and $V(x)=\sum_{n=0}^{x}2^{T_n}$. Then $$V(x)\sim\left(2^{T_{x}}+2^{T_{x-1}}\right)\prod_{n=0}^{x-2}\left(\frac{1}{1-2^{\left(T_{n}-T_{x}\right)}}\right) \tag{1}$$ Comparing with the original values given by @Claude and the approximated values from the equation given above floored down; $$V(x)=\{1,3,11,75,1099,33867,2131019,270566475,68990043211,...\}$$ $$Eq(1)=\{2,3,11,75,1099,33869,2131036,270566743,68990051601,....\}$$

This can indeed be proven using with elementary manipulations. I know that the above stated approximation is not even the best, because it is better to just sum the original function than to approximate, but turning summations to products can be good, if you want to make an asymptotic expacntions by taking logs etc. I will maybe edit, for add in more approximations to this.

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Edit 1; I also find that the error term in Eq $1$ is $\approx2^{\frac{n(n-5)+2}{2}}$ $$V(x)\sim\left(2^{T_{x}}+2^{T_{x-1}}\right)\prod_{n=0}^{x-2}\left(\frac{1}{1-2^{\left(T_{n}-T_{x}\right)}}\right)- 2^{\frac{n(n-5)+2}{2}}\tag{2}$$ I'm also very sure that the error, even though I don't have a proof and is claiming on empirical basis, is something which we can prove because it just can't be a coincidence. Here's the corrected approximated values, along with the new errors. $$Eq(2)=\{2,3,11,75,1099,33867,2131020,270566487,68990043409,35253362138273\}$$ $$Error=\{1,0,0,0,0,0,1,12,198,6230\}$$

The error is much less in comparison to Eq.$1$.

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Edit 2; By taking Eq. $1$ and taking the log, I also got a crude assymptote for $V(x)$ $$\boxed{V(x)\sim\left(2^{T_{x}}+2^{T_{x-1}}\right)\exp\left(2^{1-2x}\right)} \tag{3}$$ Or by a little manipulation; $$\boxed{V(x)\sim\left(2^{T_{x}}+2^{T_{x-1}}\right)\left\{1+2^{1-2x}\exp\left(2^{5-2x}\right)\left(\frac{2^{x}+4}{2^{x}+1}\right)\right\}}\tag{4}$$

Mapping the values approximated by Eq.$3$ & Eq.$4$; $$Eq(3)=\{14,4,11,74,1096,33858,2130960,270565634,68990017568\}$$ $$Eq(4)=\{7.8\cdot10^{14},8945,24,76,1099,33866,2131016,270566466,68990043152\}$$

Eq.$4$ can only be applicable from $x=3$ onwards, as you can see

Others; $$V(x)\sim\left(2^{T_{x}}+2^{T_{x-1}}\right)\left(1-2^{1-3x}\right)\exp\left\{\left(2^{1-2x}+8^{1-x}\right)\exp\left(2^{\left(5-2x\right)}\right)\right\}$$ $$\sim2^{T_{x-1}}\left(1+2^{x}\right)\left(1-2^{1-3x}\right)\exp\left(2^{1-2x}+8^{1-x}\right)$$