$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}}$$
A sum!
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A sum!
Evaluate the following sum:
$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}}$$
$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}}$$
Source
Imagination is much more important than knowledge.
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 !
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Re: A sum!
First of all note that $10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$ hence
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
$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}} = \frac{135}{\sqrt{10!}} = \frac{3 \sqrt{7}}{112}$$
- $10!$ is not a perfect square,
- $10!$ has $(8+1)(4+1)(2+1)(1+1) = 270$ divisors.
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
$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}} = \frac{135}{\sqrt{10!}} = \frac{3 \sqrt{7}}{112}$$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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