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## A sum!

General Mathematics
Tolaso J Kos
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### A sum!

Evaluate the following sum:

$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}}$$
Source
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Riemann
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### Re: A sum!

Tolaso J Kos wrote:Evaluate the following sum:

$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}}$$
Source

$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}} = \frac{3\sqrt{7}}{112}$$

Full solution tomorrow morning !
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
ambren
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### Re: A sum!

gist the sum and substance of an argument. 5a(1) : the result of adding numbers the sum of 5 and 7 is 12. (2) : the limit of the sum of the first n terms of an infinite series as n increases indefinitely. b : numbers to be added broadly : a problem in arithmetic.
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ambren
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### Re: A sum!

sum noun (TOTAL) ​ [ S ] the whole number or amount when two or more numbers or amounts have been added together: The sum of 13 and 8 is 21.

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Riemann
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### Re: A sum!

First of all note that $10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$ hence
1. $10!$ is not a perfect square,
2. $10!$ has $(8+1)(4+1)(2+1)(1+1) = 270$ divisors.
If $d \mid 10!$ , then there exists $p$ such that $dp =10!$ meaning that $p$ is also a divisor of $10!$. We also note that if one of $p, q$ is less than $10!$ then the other is greater than $10!$. Note that the case $p=q$ is impossible since $10!$ is not a perfect square.

Taking into account the addent that corresponds to $d$ along with the addent that corresponds to $p$ , we get:

$$\frac{1}{d+\sqrt{10!}} + \frac{1}{p+\sqrt{10!}}= \frac{1}{d+\sqrt{10!}} + \frac{1}{ \frac {10!}{d}+\sqrt{10!}} = \frac{1}{d+\sqrt{10!}} + \frac{d}{\sqrt{10!}( \sqrt{10!}+d)} = \frac{1}{\sqrt{10!}}$$

which is constant. Since we have $\frac{270}{2} = 135$ such pairs , we conclude that

$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}} = \frac{135}{\sqrt{10!}} = \frac{3 \sqrt{7}}{112}$$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$