Rachel Burns

Posted on Tuesday, 14 October, 2003 - 05:08 pm:

hi
how would you find upper and lower bounds on the sum
1/101+ 2/102 + 3/103 +...+99/199

Thanks

Mark

Posted on Tuesday, 14 October, 2003 - 06:35 pm:

99
å
r=1

(r)/(r+100)

David Loeffler

Posted on Tuesday, 14 October, 2003 - 06:42 pm:

Do you know about integration? If so, here's a clever trick for this sort of problem. Think about the integral ò₀¹ x/(1+x) dx. Divide the interval from 0 to 1 into 100 equal subintervals, and work out upper and lower bounds for the area under that bit of the graph. You should find that when you add these up both the upper and lower bounds magically turn into something closely related to your series. Now calculate the integral in the usual way, and compare this with the upper and lower bounds you've just written down.

(You should find that your sum is roughly 100 * (1-log(2)).)

David

Shu Cao

Posted on Tuesday, 14 October, 2003 - 09:03 pm:

I don't really understand. I am simply quite stupid. What exactly does the original question mean? I can see how the integration works, but what does it mean by 'upper and lower bounds for the area under that bit of the graph'.

Chris

Posted on Tuesday, 14 October, 2003 - 09:54 pm:

Upper and lower bounds are values between which the actual value that we want to find lies, hence they are the bounds on the real value, since it can lie anywhere in the interval, but it must lie in it.

Rachel Burns

Posted on Wednesday, 15 October, 2003 - 07:28 am:

What bounds would you use for the integrals of davids method?

David Loeffler

Posted on Wednesday, 15 October, 2003 - 08:12 am:

For example, in the interval from 0 to 1/100, the function is at least 0 and at most 1/100/(1+1/100). (You'll need to convince yourself it's increasing).

So ò₀^1/100 f(x) dx is between 1/100 * 0 and 1/100 * (1/100)/(1+1/100).

Now work out a general formula for the integral from i/100 to (i+1)/100, and add all these up.

David