Integration - Inequalities (1 Viewer)

[Premus]

Hey

Here's the question:

Let n be a positive integer:

a) by considering the graph of y = 1/x show that

1 / (n+1) < (integral of 1/ x dx from n to n+1 ) < 1 /n


b) hence deduce that:

( 1 + 1/n) ^ n < e < (1 + 1/n ) ^n+1

Thanks!!

 

[mojako]

(a)
Draw the graph and think of the integral as the area under the curve.
Let's recall the formula for the area of a rectangle (just do it... you'll see why later).
Area = width * height.

The width is 1 (one) because it's the area between n and n+1.
So, numerically (I mean, without considering the units),
Area = height
Now, let height = 1/x (x-coord at a particular point)

Now if the height is 1 / (n+1) then the rectangle is smaller than the area under the curve (can you visualise it? you should draw the curve if you haven't).. if the height is 1/n the rectangle is bigger...

(b)
I've been thinking about this but my thinking doesn't seem to show any progress... sorry..

 

[CM_Tutor]

Note that we have &int; (n --> n+1) 1 / x dx = [ln|x|] (n --> n+1) = ln(n + 1) - ln n = ln[(n + 1) / n] = ln(1 + 1 / n), noting that n > 0

So, our inequality is actually:

1 / (n + 1) < ln(1 + 1 / n) < 1 / n​

Taking MHS < RHS, we have

ln(1 + 1 / n) < 1 / n​

On multiplication by n, we get

n * ln(1 + 1 / n) < 1​

and on using log laws, this becomes

ln[(1 + 1 / n)ⁿ] < 1​

Converting to exponential form, we get

(1 + 1 / n)ⁿ < e¹ = e​

This is LHS < MHS of our goal inequality. If you use LHS < MHS of the original inequality, you should be able to find the other part of the desired result.

 

[CM_Tutor]

Note to Extn 2 people. It wouold not be an unreasonable 4u extension to this question to add a part (c):

(c) Hence, show that lim_n-->&infin; (1 + 1 / n)ⁿ = e