proof 2

Question

I'm a beginner in demonstrations and I'm trying to prove that $$2< \lim_{n\to\infty} \left(1+\frac1n\right)^n< 3.$$

I proved $$2< \lim_{n\to\infty} \left(1+\frac1n\right)^n,$$ by Newton's binomial, but can't prove that $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n< 3.$$

Can you give me a light?

(I know this converges to "$e$", but I want to prove it without this.)

I know this converges to "$e$", but I want to prove it without this. — Muradin Bronzebeard, Feb 22 '18 at 00:18
Typically the way you prove $e<3$ is by using $e=\sum\frac{1}{n!}$ and getting a geometric sequence larger than the sum. — Rene Schipperus, Feb 22 '18 at 00:26

user · Answer 1 · 2018-02-22T00:39:32.947

2

HINT

Use that, by Binomial theorem

$$\left(1+\frac1n\right)^n\le\sum_{k=0}^n \frac1{k!}=1+\sum_{k=1}^n \frac1{k!}\le 1+\sum_{k=1}^n \frac1{2^{k-1}}=1+\sum_{k=0}^{n-1} \frac1{2^k}$$

and note also that by geometric series

$$\sum_{k=0}^{n-1} \frac1{2^k}=\frac{1-\frac1{2^n}}{1-\frac12}=2\left(1-\frac1{2^n}\right)<2$$

edited Feb 22 '18 at 00:39

answered Feb 22 '18 at 00:15

user

154,566

I suspect that this is an exercise in limits assigned before the limit is known to be $e$. Let's see what the OP says. – Ethan Bolker Feb 22 '18 at 00:17
ops, yes you are right! – user Feb 22 '18 at 00:17
thanks for help, I got it! – Muradin Bronzebeard Feb 22 '18 at 00:43
You are welcome, Sorry but st first I didn’t realize what you were asking for. Have you tried in this way? Let me know for any doubt. Bye – user Feb 22 '18 at 01:08

score 2 · Accepted Answer · answered Feb 22 '18 at 00:33

Use $AM-GM$ first, to show that $(1 + \frac 1n)^n \leq (1 + \frac 1{n+1})^{n+1}$ for all $n \geq 1$. To do this, perform $AM-GM$ on $n$ copies of $(1 + \frac 1n)$ and the number $1$. The arithmetic mean is $\frac{1 + n(1 + \frac 1n)}{n+1} = \frac{n+2}{n+1} = 1 + \frac 1{n+1}$ and the geometric mean is $\sqrt[n+1] {(1 + \frac 1n)^n}$, so $AM \geq GM$ gives the desired result.

Since $n=1$ gives $(1 + \frac 1n)^n = 2$, we get that every term of the sequence is greater than or equal to $2$, so the limit is greater than or equal to $2$.

For the other bound, I want you to complete the steps I give you, because the proof is elementary, but it would not be good for you if I just do it and give it to you.We will prove that $(1 + \frac 1n)^n < 3$ for all $n$. This will show that the limit must also be less than (or equal to) $3$.

First, binomially expand $(1 + \frac 1n)^n$. You will get a sum, of some powers of $n$ times some binomial coefficients which depend on $n$.
Take all the negative powers of $n$ to the numerator, so you have some expression looking like $1 + \frac {c_1}{1!} + \frac{c_2}{2!} + ... + \frac{c_n}{n!}$,where $c_i$ are expressions which will depend on $n$.
See that when you do this, each $c_i \leq 1$. So you get an inequality, that $(1 + \frac 1n)^{n} \leq 1 + \frac 1{1!} + \frac 1{2!} + ... + \frac 1{n!}$.
Use the fact that $n! > 2^{n-1}$ to bound this sum above by a geometric series.
Sum the geometric series to get the result.

You are welcome! By the way, you are asking for $2 \mathbf{<} e \mathbf{<} 3$, but I could only give you $\leq$. I think a further improvement would require more tools or a refinement of the above argument. — Sarvesh Ravichandran Iyer, Feb 22 '18 at 00:46

proof 2

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