## anonymous one year ago Find the upper-bound for the summation, such that: (using the lowest possible integer)

1. amistre64

it feels like there is something missing in the post

2. anonymous

oops

3. anonymous

$\sum_{n=4200}^{x}n ^{1.5} \ge 1.0 \times 10^{8}$

4. anonymous

not sure if possible. I could solve it if it was something like 5n

5. amistre64

does this have anything to do with a remainder thrm?

6. anonymous

I want to solve it for a game. Putting in random numbers, the closest integer I could get to have the same as close as possible to 100 million is 4563

7. anonymous

*starting at 4220 >_>

8. anonymous

Wolfram gives me weird zeta functions or harmonic numbers

9. amistre64

10^8, or 10^7?

10. anonymous

100 million, so 10^8 (if I did that right)

11. amistre64

$\int_{4200}^{x}x^{1.5}dx=x^{2.5}/2.5-(4200)^{2.5}/2.5$ $x^{2.5}/2.5-(4200)^{2.5}/2.5=10^8$ $x^{2.5}-(4200)^{2.5}=2.5*10^8$ $x^{2.5}=(4200)^{2.5}+2.5*10^8$ $x=((4200)^{2.5}+2.5*10^8$)^{1/}

12. amistre64

ugh ... firefox crashed at the end of that

13. amistre64

$x=((4200)^{2.5}+2.5*10^8)^{1/2.5}=4545.ddd$

14. anonymous

I tried integrals, I must have done it wrong.

15. amistre64
16. amistre64

letting x=4545 seems to be fair

17. anonymous

oh, I see what I did wrong. Thank you :D

18. amistre64

yw