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

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.

• Are you asking for a formula for the partial sums of the sequence $1,2,8,64,1024$, or for a formula for the $n$th term of the sequence? – angryavian Jan 21 at 6:38
• partial sums, sorry if it wasn't clear. – L. McDonald Jan 21 at 7:15
• I am not sure that hypergeometric series be the proper term. Do you have a reference ? – Yves Daoust Jan 21 at 7:58
• There does not seem to be a closed-form expression. We can design different types of approximations/asymptotic expressions. But what to do depends on why you are asking the question: just "to know", or is there a true application ? – Yves Daoust Jan 21 at 9:07
• @YvesDaoust en.wikipedia.org/wiki/Generalized_hypergeometric_function: In mathematics, a generalized hypergeometric series is a power series in which the ratio of successive coefficients indexed by n is a rational function of n. – L. McDonald Jan 22 at 8:04

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 $$a_n = 2^{n(n-1)/2}, \qquad n = 0, 1, 2,\ldots$$

• Sorry if I need to clarify the question. I'm looking for a formula for the sum of $a_1 + a_2 ... a_n$ for some n – L. McDonald Jan 21 at 7:18

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 $$\{1,2,8,64,1024,32768,2097152,268435456,68719476736,35184372088832\}$$

• Is there any way to effectively compute the value? A 'condensed formula' like there is for arithmetic and geometric series. Like the $\frac{n(a_1 + a_n)}2$ for arithmetic progressions. – L. McDonald Jan 22 at 8:08
• @L. McDonald, It's an open problem. Even if it it had closed forms, it wouldn't be in terms of elementary functions. for instance take, $$\frac{1}{2^{0}}+\frac{1}{2^{1}}+\frac{1}{2^{3}}+\frac{1}{2^{6}}+\frac{1}{2^{10}}+...=\frac{\sqrt[8]{2}}{2}\vartheta_2\left(0,\frac{1}{\sqrt{2}}\right)$$Where that exotic function is the theta function. So what you're dealing with is much related to elliptic functions than hyper-geometric. – Mourad Jan 24 at 7:10
• @Mourad oh gosh, thought it would be much simpler. Maybe I'll put some work into it when I'm better at maths lol XD. Thanks! – L. McDonald Jan 24 at 9:29

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.

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.$$1$$.

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