# Proving an inequality which looks like we could use Bernoulli's inequality

How can we prove this inequality: $$\left(1+\frac{1}{n}\right)^n<3$$ What I did is: $$(1+\frac{1}{n})^n=\sum_{k=0}^{n}\binom{n}{k}1^{(n-k)}\frac{1}{n^k}=$$ $$1+\sum_{k=1}^n\binom{n}{k}\frac{1}{n^k}$$ I got stuck here, I have to prove that $\sum_{k=1}^n\binom{n}{k}\frac{1}{n^k}<2$ Could you give me a hint?

• You could use $\binom{n}{k}\frac{1}{n^k} \leqslant \frac{1}{k!}$. Nov 6 '15 at 13:21
• @DanielFischer That seems to make sense, but how could that help me?
– A6SE
Nov 6 '15 at 13:24
• You can estimate $$\sum_{k = 1}^\infty \frac{1}{k!}$$ pretty easily. Nov 6 '15 at 13:26

Look at $$L(x) = \ln f(x) = n \ln \left(1 + \frac{1}{n}\right)$$ and note that it is increasing and that $$\lim_{n \to \infty} L(x) = 1$$ (which you can show by L'Hospital's rule, for example).
Since $\ln 3 > 1$, we conclude that $L(x) \ge 1 < \ln 3$ and thus exponentiating yields $$f(x) = e^{L(x)} < e^{\ln 3} = 3$$ as desired.
• Thank you, one more question: Could you just explain me why $\sum_{k=0}^{n-1}\frac{1}{2^k}=2-\frac{1}{2^{n}}$?
• @A6Tech This about itm here is a hint: $$\frac{1}{2^k} + \frac{1}{2^k} = \frac{2}{2^k} = \frac{1}{2^{k-1}}.$$ so for example $$\frac{1}{16} + \left(\frac{1}{16} + \frac{1}{8} + \frac{1}{4} + \frac{1}{2} + 1 \right) = 2$$ If you are into doing this quickly, just sum the geometric series on the left-hand side Nov 6 '15 at 14:01