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All that we have proven so far is that limit $(1+1/n)^n$ exists and considered to be a number 'e' which belongs to $(2,3)$

We haven't proven that 'e' is irrational or that lim $(1+(x/n))^n) = e^x$

We only have the properties of sequences like Monotone convergence theorem and basic properties to prove this.

I was able to prove the previous question $((1+(1/n))^{2n})$ by using the theorem of sequences that says square of a convergent sequence converges to the square of the original limit.

But I believe that the convergence of this requires us to use $(1+(x/n))^n) = e^x$

So, if some one can prove it by basic properties of sequences, please do so!

Martin Sleziak
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Lelouch
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  • You already know it is bounded, try using $a_n > a_{n-1}$ to prove that it is monotonic and increasing , and therefore convergent. – Morgormir Feb 15 '16 at 13:10
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    Also, $$\biggl(1 - \frac{1}{n}\biggr)^n = \frac{\bigl(1 - \frac{1}{n}\bigr)^n\bigl(1+\frac{1}{n}\bigr)^n}{\bigl(1+\frac{1}{n}\bigr)^n} = \frac{\bigl(1 - \frac{1}{n^2}\bigr)^n}{\bigl(1+\frac{1}{n}\bigr)^n}.$$ And by Bernoulli's inequality $1 - \frac{n}{n^2} \leqslant \bigl(1-\frac{1}{n^2}\bigr)^n \leqslant 1$. – Daniel Fischer Feb 15 '16 at 13:18
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    @DanielFischer interesting use of Bernoulli. – Morgormir Feb 15 '16 at 13:27
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    Related posts: http://math.stackexchange.com/questions/596771/finding-the-limit-of-left-fracnn1-rightn http://math.stackexchange.com/questions/124999/how-do-i-find-lim-n-to-infty-fracn-1nn http://math.stackexchange.com/questions/269363/proof-lim-n-to-infty-1-frac1n-n-e http://math.stackexchange.com/questions/1531366/limit-of-left-1-frac1n-rightn And you can also try to have a look at other posts linked there. – Martin Sleziak Feb 15 '16 at 14:18

1 Answers1

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$$\left(1-\frac1n\right)^n = \left(\frac{n-1}n\right)^n = \frac 1{\left(\frac{n}{n-1}\right)^n} = \frac 1{\left(\frac{n-1+1}{n-1}\right)^n} = \frac 1{\left(1+\frac1{n-1}\right)^n} = \frac 1{\left(1+\frac1{n-1}\right)^{(n-1)n/(n-1)}} = \cdots $$

Martín-Blas Pérez Pinilla
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    For the last step, one can even avoid the ratio in the exponent: $(1+\frac{1}{n-1})^n = (1+\frac{1}{n-1})^{n-1}\cdot (1+\frac{1}{n-1})$, the first goes to $e$ and the second to $1$. – Clement C. Feb 15 '16 at 13:13