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Please help me understand this I know how the limit equals $e$ and $1/e$ but how is this happening I would like to see the proof

Brian Tung
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Arya Tripathi magicauras
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  • Because of this: https://math.stackexchange.com/a/579415/42969 – Martin R Nov 20 '23 at 14:46
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    See also https://math.stackexchange.com/q/1728752/42969 – Martin R Nov 20 '23 at 14:47
  • A fair amount depends on how you've defined $e$. Sometimes $e$ is defined to be that limit; other times, it's defined in some other way (such as the power series) and you have to show that the limit exists and yields the same result as that definition. In your studies, how has $e$ been defined? – Brian Tung Nov 20 '23 at 14:48
  • In my studies e hasn't even been introduced lol but thankss – Arya Tripathi magicauras Nov 20 '23 at 14:56
  • Algebraic manipulation yields

    $$\Big(1-\frac1n\Big)^n=\Big(\frac{n-1}{n}\Big)^n=\frac{1}{\Big(\frac{n}{n-1}\Big)^n}=\frac{1}{\Big(1+\frac{1}{n-1}\Big)^{n-1}\Big(1+\frac{1}{n-1}\Big)}$$

    for $n\geq2$. The term $\Big(1+\frac{1}{n-1}\Big)^{n-1}\xrightarrow{n\rightarrow\infty}e$; the term $1+\frac{1}{n-1}\xrightarrow{n\rightarrow\infty}1$. Putting things together, one gets $$\Big(1-\frac{1}{n}\Big)^n\xrightarrow{n\rightarrow\infty}\frac{1}{e}$$

     – Mittens Nov 20 '23 at 16:29

2 Answers2

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If you divide the first expression by the second one, you get $$ \frac{\left(1+\frac{1}{n}\right)^n}{1/(1-1/n)^n}=\left(1+\frac{1}{n}\right)^n \cdot \left(1-\frac{1}{n}\right)^n = \left(1-\frac{1}{n^2}\right)^n. $$ You should be able to show that this converges to $1$ as $n\rightarrow\infty$ (can you?). Which proves that the two sequences in the title must have the same limit (assuming either converges).

mjqxxxx
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Observe that $1-\frac{1}{n}=\frac{n-1}{n}$
Thus, we have $$\frac{1}{(1-1/n)^n}=(\frac{n}{n-1})^n=(1+\frac{1}{n-1})^n=(1+\frac{1}{n-1})^{n-1} \cdot \frac{n-1}{n}$$Taking $\lim_\limits{n \to \infty}(1+\frac{1}{n-1})^{n-1} \cdot \frac{n-1}{n}=\lim_\limits{n \to \infty}(1+\frac{1}{n-1})^{n-1} \cdot \lim_\limits{n \to \infty} \frac{n-1}{n}=e \cdot 1=e$

Vasili
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