I am studying computer science in first term, and i got a task that i was not able to solve for a long time now.
I have to prove that
$ 3 > (1+\frac{1}{n})^n>=2$
for every $n \in \mathbb{N}$
$(1+\frac{1}{n})^n>=2$
Can be proved easily with Bernoullis Inequality:
$(1+x)^n>=1+x*n$
$(2= 1+\frac{1}{n}*n)$
Thats cool. But how do i prove that it is smaller than 3? I thought of the following. Using bionomical theorem we can write the above term as:
$$\sum_{k=0}^n\binom{n}{k}*\frac{1}{n^k}$$
If we write the first to sums for k= 0 and k= 1 seperately, we can see that they both equal 1. $$\binom{n}{0}*\frac{1}{n^0}$$ And
$$\binom{n}{1}*\frac{1}{n^1}$$ Both equal one, am i correct.
So the sum $$\sum_{k=2}^N\binom{n}{k}*\frac{1}{n^k}$$ (Without the first two sums) has to be >1. But thats where my problem is. I mean i already proved that Each of the sums is smaller than 1 but that doesnt prove anything, right? $>1+>1\neq >1$ if you know what i mean. The whole thing has to be proofed to be smaller than 1. How can i see that.
P.S: The answer is NOT that the above term wont get bigger as Eulers Number, as by definition. I mean this statement is correct, but it would not solve the question. I really have to, and want, to show it on my own.
Any help is much apprechiated, been thinking about this for so many hours now.