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$$\lim_{n\to \infty} \left(1+\frac{1}{n!}\right)^{2n} $$

I've tried: $$\lim_{n\to \infty} \left(1+\frac{1}{n!}\right)^{2n} = \lim_{n\to \infty} e^{\ln{\left(1+\frac{1}{n!}\right)^{2n}}} = \lim_{n\to \infty} e^{2n\ \ln{\left(1+\frac{1}{n!}\right)}}$$ But I don't know how to work with the factorial

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    $\begingroup$ $1/n!$ goes to 0 so hard that it hurts. It must be 1. I wouldn't use that argument in an exam though. $\endgroup$ Mar 2 '13 at 19:50
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$$\lim_{n\to \infty} \left(1+\frac{1}{n!}\right)^{2n}=\lim_{n\to \infty}\left( \left(1+\frac{1}{n!}\right)^{n!}\right)^{\frac2{(n-1)!}}=e^0=1$$


Alternatively, Let $$A=\left(1+\frac{1}{n!}\right)^{2n}$$

So, $$\log A=2n\log \left(1+\frac{1}{n!}\right)\text { using }\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots$$

which comes from the Taylor series which holds for $|x|<1$ as $n\to\infty, \frac1{n!}\to0\implies |\lim_{n\to\infty}\frac1{n!}|<1 $

$$=2n\left(\frac1{n!}-\frac{\frac1{(n!)^2}}2+\frac{\frac1{(n!)^3}}3+\cdots\right)$$

$$=\frac2{(n-1)!}-\frac1{n!(n-1)!}+\frac{2}{3(n-1)!(n!)^2}+\cdots$$

So, $$\lim_{n\to \infty}\log A=0\implies \lim_{n\to \infty}A=e^0=1 $$

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  • $\begingroup$ the exponent is $2n$ right? why did you write $\frac{2}{n}$ when taking the logarithm? $\endgroup$
    – milo
    Mar 1 '13 at 14:34
  • $\begingroup$ @milo, rectifying $\endgroup$ Mar 1 '13 at 14:35
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    $\begingroup$ @milo, thanks for your observation. $\endgroup$ Mar 1 '13 at 14:45
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    $\begingroup$ The first line would need some more explanation, like $\forall n (1+n^{-1})^n<e$ is also true - in some similar exercises these kind of limit-taking could lead to some nasty errors. $\endgroup$ Oct 26 '15 at 7:39
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Since the exponential function is non-negative, we have for $x\geq 0$, $$0\leq \int_0^x\int_0^y \exp(z)\,dz\,dy=\exp(x)-x-1. $$ Plugging into the inequality $1+x \leq \exp(x)$ gives $$1\leq \left(1+\frac{1}{n!}\right)^{2n}\leq \exp\left(2n/n!\right)=\exp\left(2/(n-1)!\right)\to 1$$

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  • $\begingroup$ how do you prove the inequality? $\endgroup$
    – milo
    Mar 1 '13 at 14:53
  • $\begingroup$ @milo I've added a proof of the inequality. $\endgroup$
    – user940
    Mar 1 '13 at 16:52
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Hint: $$\ln(1+x)=x+O(x^2)$$ as $x\to 0$. (So this limit is equal to 1.)

or you can use inequality $$\ln(1+x)\le x.$$

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Here, I complete @milo's try: (we use $\ln(1+x)\sim_0x) $$\lim_{n\to \infty} \left(1+\frac{1}{n!}\right)^{2n} = \lim_{n\to \infty} e^{\ln{\left(1+\frac{1}{n!}\right)^{2n}}} = \lim_{n\to \infty} e^{2n\ \ln{\left(1+\frac{1}{n!}\right)}}=\lim_{n\to \infty} e^{\frac{2n}{n!}}=e^0=1.$$

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