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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?

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

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  • $\begingroup$ Thank you, one more question: Could you just explain me why $\sum_{k=0}^{n-1}\frac{1}{2^k}=2-\frac{1}{2^{n}}$? $\endgroup$
    – A6SE
    Nov 6 '15 at 13:43
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    $\begingroup$ @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 $\endgroup$
    – gt6989b
    Nov 6 '15 at 14:01

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