FP2 help

I'm stuck on question number 7. http://www.ocr.org.uk/Images/136137-question-paper-unit-4726-further-pure-mathematics-2.pdf

I've done part 1 a=2 b=n c=1 d=n-1

I'm not really too sure how to do part 2. I've integrated y=1/x using the limits from 1 to n, to get ln(n), but then I'm not too sure what to do.

Substituting those values in we have

$\displaystyle \frac{1}{2}+\frac{1}{3}+...+ \frac{1}{n}<\ln n < 1+ \frac{1}{2}+...+\frac{1}{n-1}$


I'll start by finding the upper bound:

$\displaystyle f(n)=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n$


We have a lower bound for $\ln n$ so replacing $\ln n$ with that lower bound will make $f(n)$ bigger:

$\displaystyle f(n)=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n < 1+\frac{1}{2}+ \frac{1}{3}+...+ \frac{1}{n} - \left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)$

Can you use this to find the upper bound? Then follow a similar procedure to find the lower bound.

So everything apart from the 1 on the RHS cancels so we get f(n)<1 so the upper bound is 1? Also if that is correct can you explain why this is the upperbound I would have thought it is the lower bound... (Sorry if I'm being stupid).

So everything apart from the 1 on the RHS cancels so we get f(n)<1 so the upper bound is 1? Also if that is correct can you explain why this is the upperbound I would have thought it is the lower bound... (Sorry if I'm being stupid).

1 is correct.

If you're looking to make something as large as possible and subtracting something, you want the number you're subtracting to be as small as possible.

e.g.

2<3<4

5-4 < 5-3 < 5-2

1 is correct.

If you're looking to make something as large as possible and subtracting something, you want the number you're subtracting to be as small as possible.

e.g.

2<3<4

5-4 < 5-3 < 5-2

So i reattempted the question and I've confused myself.

the integral comes out as ln(n) so we can rewrite the inequalities as

$\frac {1}{2}+\frac{1}{3}+...+\frac{1}{n}<ln(n)<1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n-1}$. So then we can rearrnage to get

$\frac {1}{2}+\frac{1}{3}+...+\frac{1}{n}-ln(n)<1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n-1}$. Which is the same as

$f(n)<1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n-1}$. From here I get stuck.

If you're going to subtract $\ln n$ from one side of the inequality, you need to subtract it from all other sides as well:

$\displaystyle \frac {1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n<\ln n-\ln n<1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n-1} - \ln n$

Try continuing from here.

So we get $f(n)<0<f(n-1)$? I'm not too sure on the RHS.

The LHS is not f(n), it's missing a 1. And the RHS is missing a 1/n term from f(n). So

$\displaystyle \frac {1}{2}+\frac{1}{3}+...+\frac{1}{ n}-\ln <\ln n-\ln n<1+\frac{1}{2}+\frac{1}{3}+...+ \frac{1}{n-1} - \ln n$

$\displaystyle \implies f(n)-1<0<f(n)-\frac{1}{n}$


I assume because you are using this method, you weren't comfortable with the method I gave?

I initially thought I was, but when i tried it I got really confused, I'll look over it again though.

I got the same as notnek when I did it, which gives the bounds 1/n < f(n) < 1, although I actually thought I was doing the same as him but in 1 stage rather than 2, as opposed to giving you a completely different method