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Why does $$\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n=e$$ but $$\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n=e^{-1}$$

Shouldn't the limits be the same since $\left(1+\frac{1}{n}\right) \to 1$?

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    $\begingroup$ No............. $\endgroup$ Dec 2 '17 at 20:11
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    $\begingroup$ $\lim_{n \to \infty} \left(1+\frac an\right)^n = e^a$, for $a \in \mathbb{R}$. $\endgroup$
    – TheSimpliFire
    Dec 2 '17 at 20:12
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    $\begingroup$ Generally speaking, $1^{+\infty}$ is an indeterminate form. The limit can be anything. $\endgroup$ Dec 2 '17 at 20:14
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    $\begingroup$ @amWhy I think there is a reason given for why the two limits should be the same. It is not a good reason and exposes a misunderstanding, and I think it is clear that OP is trying to understand why their intuition isn't working. $\endgroup$ Dec 2 '17 at 20:52
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    $\begingroup$ Oh, sure, @MarkBennet, now that you've edited the question to improve its quality. My comment was left prior to your edit, and was valid wrt the original answer. Don't chastise someone for observing a problem with the post, after you have edited it to improve it for the asker. $\endgroup$
    – amWhy
    Dec 2 '17 at 21:04
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If the limits were the same, they would have to be $1$, since $$\left(1+\frac1n\right)^n>1$$ for all $n$, while $$\left(1-\frac1n\right)^n<1$$ for all $n$.

So if you expect the first limit to be $e$ and you know that $e>1$, you cannot expect the second one to be equal to it.

Edit: (to incorporate two of Paramanad Singh comments). One can obtain from Bernoulli's Inequality that $$\left(1+\frac1n\right)^n\geq2,\ \ n\geq1.$$ Which implies that $e\geq2$ as soon as one knows that the limits exists. This already shows that the limits cannot be the same. More interestingly, from $$ \left(1-\frac{1}{n}\right)^{n}=\dfrac{1}{\left(1+\dfrac{1}{n-1}\right)^{n-1}}\cdot\left(1-\frac{1} {n}\right),n>1 $$ one gets that $$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^{n}=e^{-1}.$$

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    $\begingroup$ (+1) very simple explanation of why the limits are different. $\endgroup$
    – zwim
    Dec 2 '17 at 20:51
  • $\begingroup$ (+1) More interestingly you can show that expression $(1+(1/n))^{n}$ is always greater than $2$ (via Bernoulli inequality ) and the expression $(1-(1/n))^{n}$ is trivially less than $1$ so that there is no way they can tend to the same limit. This deals with the part "you know that $e>1$" of your answer. $\endgroup$
    – Paramanand Singh
    Dec 3 '17 at 3:22
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It's the same reason that you can't argue that $\lim_{n \to \infty}(1 + \frac{1}{n})^n = 1$, even though $1 + \frac{1}{n} \to 1$ and $1^n = 1$. The issue is that when we say that the limit of an expression is a certain value, we just mean the expression gets very close to that value - we make no promises about how fast.

When $n$ is very large, $1 - \frac{1}{n}$ is very nearly $1$ - but a number very nearly $1$, when raised to a very large power, can be very small. For example, $0.99^{10000}$ is zero to more than forty decimal places. So that exponent of $n$ can "pull" the value away from one - for $(1 - \frac{1}{n})^n$, it pulls it down, while it pulls $(1 + \frac{1}{n})^n$ up.

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Just to give you some intuition, note that since $1-x < 1/(1+x)$ for $0<x<1,$

$$ (1-1/n)^n < \frac{1}{(1 + 1/n)^n}$$

for $n>1.$ Now the right side $\to 1/e.$ Thus if the limit of the left side exists, it has to be $\le 1/e.$

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Expand both using the binomial theorem

$$\left(1+\frac 1n\right)^n=1^n+n\cdot\frac 1n\cdot1^{n-1}+\binom n2\left(\frac 1n\right)^21^{n-2}+\binom n3\left(\frac 1n\right)^31^{n-3}+\dots =$$$$=1+1+\frac {n(n-1)}{2n^2}+\frac {n(n-1)(n-2)}{6n^3}+\dots$$while $$\left(1-\frac 1n\right)^n=1-1+\frac {n(n-1)}{2n^2}-\frac {n(n-1)(n-2)}{6n^3}+\dots$$

Subtract the second from the first and you get $$2+\frac {n(n-1)(n-2)}{3n^3}+\dots$$ a sum of positive terms. So the difference between the two expressions is at least $2$. If the limits exist (they do), the difference between the limits must be at least $2$.

If you analyse the difference you will note that it is increasing with $n$ (but bounded) so even though the $(1\pm \frac 1n)$ part tends to zero, raising to the $n^{th}$ power in this case gives expressions which get further apart rather than closer together.

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