Divisor Sigma Function

Calculus Level 5

$S=\displaystyle \sum_{p} \displaystyle \sum_{n=2}^{\infty} \dfrac{\tau (n)-1}{p^n}$

where $p\in\mathbb{N}$ is greater than 1 and is NOT a perfect power.

Find $\lceil 10000S \rceil$

Notes:

$\tau (x)$ is the divisor sigma function. This represents the number of factors of a number including 1 and itself. For example, $\tau(6)=\sigma_0(6)=4$ .
$\lceil \cdot \rceil$ denotes the ceiling function .

The answer is 10000.

1 solution

Trevor Arashiro
Oct 27, 2016

This is the cool (but hard) way to do this problem.

$\zeta(1)=\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....$

$\displaystyle \sum_1^{\infty} \dfrac{1}{2^x}=1$

$\displaystyle \sum_1^{\infty} \dfrac{1}{3^x}=\dfrac{1}{2}$

$\displaystyle \sum_1^{\infty} \dfrac{1}{4^x}=\dfrac{1}{3}$

$(i)~\zeta(1)=\displaystyle \sum_1^{\infty} \dfrac{1}{2^x}+\displaystyle \sum_1^{\infty} \dfrac{1}{3^x}+\displaystyle \sum_1^{\infty} \dfrac{1}{4^x}+\displaystyle \sum_1^{\infty} \dfrac{1}{5^x}+...$

$\begin{aligned} \zeta(1)&=&\dfrac{1}{1}+&\dfrac{1}{2}+&\dfrac{1}{3}+&\dfrac{1}{4}+...\\ --&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&\\ \\ \zeta(1)&=&&\dfrac{1}{2}+&&\dfrac{1}{4}+&&\dfrac{1}{8}+&&&&\dfrac{1}{16}+&&&\dfrac{1}{32}+&\dfrac{1}{64}+....\\ &&&&\dfrac{1}{3}+&&&&\dfrac{1}{9}+&&&&&\dfrac{1}{27}+&&&\dfrac{1}{81}+....\\ &&&&&\dfrac{1}{4}+&&&&&&\dfrac{1}{16}+&&&&\dfrac{1}{64}+....\\ &&&&&&\dfrac{1}{5}+&&&&&&\dfrac{1}{25}+&&&&&\dfrac{1}{125}+....\\ \\ \end{aligned}$

Each row represents the expanded sum of $\displaystyle \sum_{x=1}^{\infty} \dfrac{1}{a^x}$ for some $a$

Note that the first diagonal is $\zeta(1)-1$ and that the sum of all the diagonals is the same as line $(i)$ . Since the sum of all the diagonals is $\zeta(1)$ , the sum of all the diagonals after the first diagonal must be 1. Each term will appear the same number of times as the number of factors of its power minus one (minus one because each term is included once in in the first diagonal, $\zeta(1)$ ). We can observe this visually by summing the terms.

$1=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{2}{16}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{32}+\dfrac{1}{36}+\dfrac{1}{49}+\dfrac{3}{64}+\dfrac{2}{81}+...$

Hence

$1=\displaystyle \sum_{p} \displaystyle \sum_{n=2}^{\infty} \dfrac{\tau (n)-1}{p^n}$

The simpler way to solve this problem is by rearranging the sum

$\begin{aligned} \displaystyle \sum_{p} \displaystyle \sum_{n=2}^{\infty} \dfrac{\tau (n)-1}{p^n}&=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{32}+\dfrac{1}{36}+\dfrac{1}{49}+\dfrac{1}{64}+\dfrac{1}{64}+\dfrac{1}{64}+\dfrac{1}{81}+\dfrac{1}{81}+... \\ &=\left(\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+....\right)+\left(\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...\right)+ \left(\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+...\right)+\left(\dfrac{1}{25}+\dfrac{1}{125}+...\right)\\ &=\displaystyle \sum_2^{\infty} \dfrac{1}{2^x}+\displaystyle \sum_2^{\infty} \dfrac{1}{3^x}+\displaystyle \sum_2^{\infty} \dfrac{1}{4^x}+\displaystyle \sum_2^{\infty} \dfrac{1}{5^x}+...\\ &=\displaystyle \sum_{a=2}^{\infty} \displaystyle \sum_{x=2}^{\infty} \dfrac{1}{a^x}\\ &=\displaystyle \sum_{a=2}^{\infty} \dfrac{1}{a(a-1)}\\ &=\displaystyle \sum_{a=2}^{\infty} \dfrac{1}{(a-1)}-\dfrac{1}{a}\\ &=1 \end{aligned}$

I think you are trying to replicate the following result: $\sum_{m = 2}^\infty \sum_{k = 2}^\infty \frac{1}{m^k} = 1.$ In other words, the sum of the reciprocals of the perfect powers (excluding 1) is 1, where each perfect power is counted as many times it can be represented as a perfect power (such as $\frac{1}{64} = \frac{1}{2^6} = \frac{1}{4^3} = \frac{1}{8^2}$ ).

However, this is not what your sum counts. Again, take the perfect power $64 = 2^6$ . For $p = 64$ and $n = 6$ , $\frac{\tau(n) - 1}{p^n} = \frac{3}{64^6},$ which is not equal to $\frac{1}{64} + \frac{1}{64} + \frac{1}{64}$ .

(Also, the first part of your argument has a serious flaw. You argue that your sum is equal to $\zeta(1) - [\zeta(1) - 1]$ , but $\zeta(1)$ diverges, so your sum is of the indeterminate form $\infty - \infty$ , which is meaningless.)

Jon Haussmann - 4 years, 7 months ago

Sorry, the problem was wrongly edited. You have now been marked as correct.

@Trevor Arashiro is fixing the problem.

Calvin Lin Staff - 4 years, 7 months ago

Sorry about that. While trying to make the problem look nicer, one of the mods accidentally changed the problem's meaning. I have changed the problem to its original meaning.

Trevor Arashiro - 4 years, 7 months ago

Divisor Sigma Function

The answer is 10000.

1 solution

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