# Estimating e

1. Feb 19, 2005

### matrix_204

how do i show that 1/(n+1)<=ln(1+1/n)<=1/n? i need sum help, i have the definitions for log n e except i need to kno how to apply them, at first it seemed simple to just do it but i m havin sum trouble..

2. Feb 19, 2005

### nolachrymose

I'm not quite sure if this is correct, but some things I noticed might be of help:

1. Well, one part of the inequality is trivial. It is quite clear that 1/(n+1) <= 1/n.
2. If you multiply all sides by n (and apply the logarithmic property to the middle term), you will find the inequality easier, most likely.
3. Try applying the "squeeze principle."

Hope that helps! :)

Last edited: Feb 19, 2005
3. Feb 19, 2005

### matrix_204

what's the squeeze principle, is that the pinching theorem?
also the reason i asked this question was cuz like u mentioned, it's kind of obvious but i was wondering what else it might involve besides just knowing that 1/(n+1) <= 1/n.

4. Feb 19, 2005

### Curious3141

Try this :

Sketch the curve $$y = \frac{1}{1 + x}$$ for the domain of $$x \geq 0$$

Now demarcate the area under the curve bounded by the vertical lines $$x = 0$$ and $$x = \frac{1}{n}$$

Find the ordinates (y-values) at those x-values. Come up with two rectangles that form upper and lower bounds for the area and work out their areas in terms of n.

Now evaluate the same area exactly using definite integration. What can you now see ?

5. Feb 20, 2005

### xanthym

You've already solved this problem!! Let x=(1 + 1/n) and go here:

$$:(1): \ \ \ \ \frac {x - 1} {x} \ \leq \ ln(x) \ \leq \ (x - 1)$$

$$:(2): \ \ \ \ x = 1 + \frac {1} {n}$$

$$:(3): \ \ \ \ \frac {1} {n + 1} \ \leq \ ln(1 + \frac {1} {n}) \ \leq \ \frac {1} {n}$$

~~

6. Feb 20, 2005

### matrix_204

oooh i didn't even notice that, ok so this one i have done it, now there is another similar type of problem, except this one involves e, it goes like this;
(1+1/n)^n<= e <=(1+1/n)^n+1
do i solve this one by finding the logs of all these functions, since its a power of n and so forth..(n+1)