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 Post subject: A sum!
PostPosted: Mon Nov 09, 2015 5:18 pm 
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Evaluate the following sum:

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

Source
Matha's notes.

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 Post subject: Re: A sum!
PostPosted: Wed Aug 01, 2018 9:35 am 

Joined: Sat Nov 14, 2015 6:32 am
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Location: Melbourne, Australia
Tolaso J Kos wrote:
Evaluate the following sum:

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

Source
Matha's notes.



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

Full solution tomorrow morning !

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$


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 Post subject: Re: A sum!
PostPosted: Sat Jan 12, 2019 10:45 am 

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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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 Post subject: Re: A sum!
PostPosted: Tue Jan 15, 2019 7:55 am 

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sum noun (TOTAL) ‚Äč [ 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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 Post subject: Re: A sum!
PostPosted: Thu Feb 07, 2019 4:57 pm 

Joined: Sat Nov 14, 2015 6:32 am
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Location: Melbourne, Australia
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

$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}} = \frac{135}{\sqrt{10!}} = \frac{3 \sqrt{7}}{112}$$

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