Matematické Fórum

stuart clark · 30. 10. 2018 15:30

A sequence $b_{n}$ given as $\displaystyle b_{1}=\frac{1}{3}$ and $\displaystyle b_{n+1}=b^2_{n}+b_{n}$ .

If $\displaystyle \mathcal{S} = \sum^{2008}_{r=2}\frac{1}{b_{r}}$ . Then $\lfloor \mathcal{S} \rfloor$ is

check_drummer · 30. 10. 2018 21:20

Hi, try partial fractions.

stuart clark · 01. 11. 2018 11:26

Thanks Check_drummer.please explain me . I am not getting

check_drummer · 02. 11. 2018 15:35

↑ stuart clark:
$\frac{1}{b^2_{n}+b_{n}}=\frac{1}{b_{n}}-\frac{1}{b_{n}+1}$

stuart clark · 05. 11. 2018 02:29

Thanks check_drummer. but from here we have calculate $\displaystyle \sum^{2008}_{r=2}\frac{1}{1+b_{r}}.$

But how can i find $\displaystyle \sum^{2008}_{r=2}\frac{1}{b_{r}}.$ Thanks

check_drummer · 05. 11. 2018 15:32

↑ stuart clark:
It is telescoping sum, you only need $1/(b_{2007}+1)$ but hopefully it can be neglected because you need only the whole part of S. But this should be proved...

Pavel · 05. 11. 2018 16:30

↑ check_drummer:

Is it realy telescoping sum? The sum

$\sum_{n=2}^{2008}\left(\frac 1{b_n}-\frac 1{b_n+1}\right)$

is different from the sum

$\sum_{n=2}^{2008}\left(\frac 1{b_n}-\frac 1{b_{n+1}}\right)$

Observation:

Autor skryl část textu. Zobrazit/skrýt!

Pavel · 06. 11. 2018 02:01

↑ stuart clark:

Solution:

Autor skryl část textu. Zobrazit/skrýt!

check_drummer · 06. 11. 2018 22:31

Pavel napsal(a):
↑ check_drummer:

Is it realy telescoping sum? The sum

$\sum_{n=2}^{2008}\left(\frac 1{b_n}-\frac 1{b_n+1}\right)$

is different from the sum

$\sum_{n=2}^{2008}\left(\frac 1{b_n}-\frac 1{b_{n+1}}\right)$

Of course... Thank you for the comment.

Matematické Fórum

#1 30. 10. 2018 15:30

series

#2 30. 10. 2018 21:20

Re: series

#3 01. 11. 2018 11:26

Re: series

#4 02. 11. 2018 15:35

Re: series

#5 05. 11. 2018 02:29

Re: series

#6 05. 11. 2018 15:32

Re: series

#7 05. 11. 2018 16:30

Re: series

#8 06. 11. 2018 02:01

Re: series

#9 06. 11. 2018 22:31

Re: series

Pavel napsal(a):

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