Matematické Fórum

stuart clark · 16. 04. 2016 04:48

If $\displaystyle k_{r} = \frac{1}{r}\;, x\in \mathbb{N}.$ Then value of $\big\lfloor \sum^{11}_{r=1}k_{r} \big\rfloor .$

where $\lfloor x \rfloor$ Represent floor function of $x$ .

check_drummer · 17. 04. 2016 22:21

↑ stuart clark:
Hi,
1) what does x mean in $\displaystyle k_{r} = \frac{1}{r}\;, x\in \mathbb{N}.$ ?
2) Why not to calculate it by definition? 11 terms is not too much...

stuart clark · 21. 04. 2016 07:19

Thanks ↑ check_drummer: i also solved it using division.

Can we solve it using DEfinite Integral or any other inequality, Thanks

Matematické Fórum

#1 16. 04. 2016 04:48

Floor sum

#2 17. 04. 2016 22:21

Re: Floor sum

#3 21. 04. 2016 07:19

Re: Floor sum

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