Need a hint for calculating the integer part of

[holezch]

Homework Statement

Calculate the integer part ( whole number part ) of

1 + 1/sqrt ( 2 ) + 1 / sqrt ( 3 ) + 1 / sqrt ( 4 ) + 1 / sqrt ( 5 ) + ... + 1 / sqrt ( 1 000 000 )


Any hints? Thanks, I'd appreciate the help

 

[dacruick]

What is a whole number part?

 

[CompuChip]

I was thinking, that you can get a lower bound from
[tex]\int_{1}^{1000000} \frac{1}{\sqrt{x}} \, dx < \sum_{i = 1}^{1000000} \frac{1}{\sqrt{i}}[/tex]
Can't think of an appropriate upper bound estimation though.

 

[holezch]

Hm, if we go for an estimation, we will have to be sure that our estimation is an integer estimate z that is z <= v or z >= v if v is the actual value and z is the closest integer? I didn't think of coming up with bounds because I thought that the answer would need some sort of precision
thanks

 

[holezch]

What is a whole number part?

the integer part

 

[holezch]

anyone have any more ideas? :(

 

[Dick]

anyone have any more ideas? :(

It's pretty much what CompuChip gestured at in post 2, though being careless about being exact. You need to bound the sum by integral estimates. Draw a bunch of rectangles representing the sum as though you were doing an upper Riemann sum of the integral of 1/sqrt(n) on the interval [1,10^6+1] with rectangles of width 1. Since it's an upper sum, the integral gives you a lower bound, right? To get an upper bound, use the function 1/sqrt(n-1). I know that goes to infinity at n=1, so you'd better use a different approximation between the first two points. Be more careful of the endpoints than CompuChip was, ok? I tried it and I THINK I got estimates close enough to get the integer part of the sum. See if you think so.

 

[holezch]

thanks, I got the answer to be 1998 and I believe that it is correct :)
I initially used 1 to a million for both end points, on 1/sqrt ( x ) and 1 / sqrt ( x - 1 ) integrals, but then I couldn't integrate 1 / sqrt ( x - 1 ) with that end point, so I started from 2 instead of 1. 1/ sqrt ( 1 ) = 1 which is an integer anyway, so I could exclude that from my approximation.
Then I got
2sqrt(million ) - 2*sqrt ( 2 ) < original sum < 2*sqrt( million -1 ) - 2. Then showed that the difference between both bounds is < 1 but > 0 , which means they are closer than adjacent integers ( which means if the original sum is in between that, I will indeed get the integer part by finding the integer part of either bound ). Well, I found the integer part of either bound and added the 1 back in of course
I had to use a calculator though :S, to calculate 2*sqrt( 1 000 000 - 1 ) - 2 , is that still fair game?

 

[Dick]

thanks, I got the answer to be 1998 and I believe that it is correct :)
I initially used 1 to a million for both end points, on 1/sqrt ( x ) and 1 / sqrt ( x - 1 ) integrals, but then I couldn't integrate 1 / sqrt ( x - 1 ) with that end point, so I started from 2 instead of 1. 1/ sqrt ( 1 ) = 1 which is an integer anyway, so I could exclude that from my approximation.
Then I got
2sqrt(million ) - 2*sqrt ( 2 ) < original sum < 2*sqrt( million -1 ) - 2. Then showed that the difference between both bounds is < 1 but > 0 , which means they are closer than adjacent integers ( which means if the original sum is in between that, I will indeed get the integer part by finding the integer part of either bound ). Well, I found the integer part of either bound and added the 1 back in of course
I had to use a calculator though :S, to calculate 2*sqrt( 1 000 000 - 1 ) - 2 , is that still fair game?

That sounds right. Sure, I would say you could use a calculator to find the value of the integral. Though you could do it without the calculator if you really HAD to, right? The point is just to do it without summing the whole series.

 

[CompuChip]

sqrt(1 000 000) = 1 000
So sqrt(1 000 000 - 1) is a little (but certainly not more than 0.5) smaller (since the square root function is monotonous) and therefore
2 * sqrt(1 000 000 - 1) is just a little smaller (but certainly not more than 1) than 2 * 1000 = 2000.
So
[tex]1997 < 2 \sqrt{1000000 - 1} - 2 < 1998[/tex].