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Let $x$ be a positive integer. I have the following iterated sums:

$$f(x)=\sum_{i_{u-1}=1}^{x}\sum_{i_{u-2}=1}^{i_{u-1}} \cdots \sum_{i_2=1}^{i_3} \sum_{i_1=1}^{i_2}i_1$$

What is the asymptotic behavior of the total?

Thank you

2 Answers 2

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\begin{eqnarray*} \sum_{i_1=1}^{i_2}i_1 & =\frac{i_2(i_2+1)}{2!} \\ \sum_{i_2=1}^{i_3} \sum_{i_1=1}^{i_2}i_1 & =\frac{i_3(i_3+1)(i_3+2)}{3!} \\ \vdots & \sum_{i_u-1=1}^{x}\cdots \sum_{i_2=1}^{i_3} \sum_{i_1=1}^{i_2}i_1 & =\frac{x(x+1) \cdots (x+u-1)}{u!} \end{eqnarray*} So the asymtopic behaviour is $x^u/u!$.

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    Thank you very much for your answer. – 2017-02-18
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You can un-expand the inside by noticing

$$i_1=\sum_{k=1}^{i_1}1$$

Which gives the combinatorial sum:

$$f_u(x)=\binom{x+u-1}u$$

Think of how many ways you can count...

This then gives the following asymptote:

$$f_u(x)\sim\frac1{u!}x^u$$

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    Thank you for your answer. Is the combinatorial sum $S$ equal to the iterated sum $f(x)$? Thanks again. – 2017-02-18
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    Ah yes it is, just subconsciously used to using $S$ fir sum. :D – 2017-02-18
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    Are you sure this binomial coeff is equal to $f_u(x)$? Because your binomial coef is equal to $\frac{(x+u)(x+u-1) \cdots (x+1)}{u!}$. It seems there is a problem. – 2017-02-18
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    I am pretty sure. Could you elaborate? – 2017-02-18
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    There is a little difference between $(x+u) \cdots (x+1)$ and $(x+u-1)\cdots x$, according to the solution from Donald. – 2017-02-18
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    Ah, whoops. Minus one on the top of the binomial coefficient. – 2017-02-18
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    Thanks, but there is still the problem with $x$. – 2017-02-18
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    :-/ now what did I do? – 2017-02-18
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    That' OK sorry. And thank you :-) – 2017-02-18
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    Ok then. Thanks for all the catches ;) – 2017-02-18
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    I think there is a problem with your solution. If I take the example of $u=3$ and $i_3=5$, I find $69$, different from $35$ by the binomial coeff. – 2017-02-27