problem involving floor sum (1 Viewer)

juantheron · May 6, 2017

$\displaystyle \bigg\lfloor\frac{2017!}{1!+2!+3!+\cdots \cdots +2016!}\bigg\rfloor$

$where $\lfloor x \rfloor $ is a floor of $x$

$For upper bound $1!+2!+3!+\cdots \cdots +2016!=2016(1!+2!+3!+\cdots +2016!)$

$2016(1!+2!+3!+\cdots \cdots +2016!)>2016(2016!+2015!)=2017!$

$\displaystyle \frac{2017!}{1!+2!+3!+\cdots \cdots +2016!}<2016$

But did not understand how to get lower bound without induction, Thanks

Kingom · May 6, 2017

Mahan1 · May 7, 2017

juantheron said:
$\displaystyle \bigg\lfloor\frac{2017!}{1!+2!+3!+\cdots \cdots +2016!}\bigg\rfloor$

$where $\lfloor x \rfloor $ is a floor of $x$

$For upper bound $1!+2!+3!+\cdots \cdots +2016!=2016(1!+2!+3!+\cdots +2016!)$

$2016(1!+2!+3!+\cdots \cdots +2016!)>2016(2016!+2015!)=2017!$

$\displaystyle \frac{2017!}{1!+2!+3!+\cdots \cdots +2016!}<2016$

But did not understand how to get lower bound without induction, Thanks

$\textrm{To find the lower bound you need to use the following inequality:}$
$\sum_{k=1}^{2016} k! = 2016! (1 + \frac{1}{2016} + \frac{1}{2016 \times 2015} + \cdots + \frac{1}{2016 \times 2015 \times 2014 \times \cdots \times 1})$

$< 2016!(1 + \frac{1}{2016}+ \frac{1}{2016 \times 2015}+ \frac{1}{2016 \times 2015}+\cdots + \frac{1}{2016 \times 2015})$

$< 2016! (1 + \frac{1}{2015}+ \frac{1}{2016 \times 2015} + \frac{1}{2016 \times 2015} + \cdots + \frac{1}{2016 \times 2015 })$

$= 2016! (1 + \frac{1}{2015} + \frac{2014}{2016 \times 2015}) < 2016! (1 + \frac{1}{2015}+ \frac{1}{2016})$

$< 2016!(1+ \frac{1}{2015} + \frac{1}{2015}) = 2016!(1+ \frac{2}{2015}) = 2016! \frac{2017}{2015}$

$\textrm{That implies } \frac{2017!}{1!+2!+\cdots + 2016!} = 2017! \frac{1}{2016!(1+\frac{1}{2016}+ \cdots + \frac{1}{2016 \times 2015 \times \cdots \times 1})}$

$= \frac{2017}{1+\frac{1}{2016}+ \cdots + \frac{1}{2016 \times 2015 \times \cdots \times 1}} > \frac{2017}{\frac{2017}{2015}} = 2015$

$\textrm{therefore} \frac{2017!}{1!+\cdots + 2016!} > 2015$

$\textrm{Hence,} \left \lfloor \frac{2017!}{1!+\cdots + 2016!} \right \rfloor \geq 2015 \ \textrm{but as you pointed out the floor value is less than 2016, therefore it is} \ 2015$

problem involving floor sum (1 Viewer)

juantheron

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Kingom

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Mahan1

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