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# Help!! Past paper question [n*ln(1+(1/n)) <= 1] ?? watch

1. Help me out!!

Im doing the IB and having my IB Maths Higher exams this friday (P2).
This is a question from May 2002 Paper 2 Option: Analysis and Approximation.

Using mean value theorem show that for all positive integers n,
n*ln(1+(1/n)) <= 1

Apparently the solution should be along these lines:
n*ln(1+(1/n)) = n*Intergral(lower=1, upper=1+(1/n)) (1/t) dt
=n (1/c) with 1<c<1+(1/n)
hence (1/c)<1 so that n*ln(1+(1/n))<1
(Hope my messy description of the solution makes sense)

Where do the lower and upper bounds for the integral come?
Any explanation to get me back on track would be needed!!
2. the integral comes from the integral defenition of the natural logarithm:

ln(x) = integral{lower=1,upper=x} 1/t dt.
3. By the mean value theorem, there is a pt c in the interval [a,b] (see fig(1)) such that the gradient at c, f'(c), is equal to the gradient of the chord joining f(a) and f(b).

Taking the curve f=1/t, we can evaluate the area under the curve, from t=a to t=b, as A=ln(b) - ln(a) (Put a=1 and b=1+1/n to get your expression).

Now looking at Fig(2), we can see that the area under the curve lies between the area of the two trapeziums, abde and abgh. You should now be able to get expressions for these areas in terms of n. (use the definition of MVT to help you get expressions/relationships) Now use inequalities between all three areas to derive the inequality you're after, viz.
n*ln(1+1/n) < 1

Mind you, I found it a bit troublesome getting the smaller area (the bigger one was easy), so maybe someone else has a quicker idea.

HTH

Edit: uh, sorry about the quality of the sketches.
Attached Images

4. Thanks a lot for the responses!
Don't know how I've missed the integral definition of the natural logarithm.
Maybe it was one of those Monday mornings..

I had a second look at the problem.

n * ln(1+(1/n)) = n * intg{lwr=1,upr=1+(1/n)} (1/t) dt

With the integral definition of MVT:
intg{lwr=a,upr=b} f(x) dx = f(c)(b-a)
We get..
intg{lwr=1,upr=1+(1/n)} (1/t) dt = (1/c)(1/n)
n cancels..
n * intg{lwr=1,upr=1+(1/n)} (1/t) dt = n (1/c) (1/n)
n * intg{lwr=1,upr=1+(1/n)} (1/t) dt = (1/c)

Since 1<c<1+(1/n) we have that (1/c) < 1
Therefore n*ln(1+(1/n)) < 1

Then a friend of mine suggested also another solution..
The property of Riemann integration states that if
f(x) and g(x) are integrable and if f(x) <= g(x) on [a,b], then
intg{lwr=a,upr=b} f(t) dt <= intg{lwr=a,upr=b} g(t) dt
We get n*ln(1+(1/n)) <= intg{lwr=1,upr=1+(1/n)} dt <= 1

Yeh, higher maths paper 1 exam starts in 173 minutes.
Can't wait.

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