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Why does $$(1+1/k)^{k}\rightarrow e\hspace{0.5cm}\text{as}\hspace{0.5cm}k\rightarrow\infty$$

why does it not approach

$$(1+1/k)^{k}\rightarrow (1+0)^k=1\hspace{0.5cm}\text{as}\hspace{0.5cm}k\rightarrow\infty$$

Mike Pierce
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helpmeh
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    This is just true by definition, that is to say this is one of several equivalent definitions of the number $e$ – cpiegore Apr 03 '16 at 18:30
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    For a fixed finite $k$, expand $(1 + \frac{1}{k})^k$. – jim Apr 03 '16 at 18:30
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    Presumably the question is then: if you define $e$ this way, why is $e \neq 1$? – Ian Apr 03 '16 at 18:30
  • I do not like the fallback of explaining things "by definition", but thanks for trying cpiegore. It seems like jim is going somewhere, but i cant see how expanding for a fixed finite k would help explain, we would just get $(1)(1)(1)...$ Ian hit the nail on the head. – helpmeh Apr 03 '16 at 18:32
  • We could observe that it is more like ${(1+\epsilon)}^k$ for a small $\epsilon <0$. If we take a number which is a little bit greater then $1$ to the $k$-th power then the result is not 1. – K. Hoffmann Apr 03 '16 at 18:34
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    @helpmeh Your second statement is not correct. You need to substitute $\infty$ for EVERY use of $k$. If you do this you get $1^\infty$ which is indeterminate. – cpiegore Apr 03 '16 at 18:34
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    @K.Hoffmann I think you should actually have $\epsilon > 0$. If $\epsilon < 0$ than the answer is $0$ – cpiegore Apr 03 '16 at 19:03
  • @cpiegore thank you for the correction here – K. Hoffmann Apr 03 '16 at 19:07
  • You can't separate limits in pieces like that. Otherwise $(1+1/k)^k\rightarrow (1+1/k)^{\infty}\rightarrow \infty$. As for why it equals e, many will say it is the definition of e. But that raises the question why is e the the base in which the derivative of $e^x$ is itself or why the anti derivative of 1/x is ln, both of which are alternative definitions. The results are covered in any calculus texts. – fleablood Apr 03 '16 at 21:31
  • It is the definition of e, (or one at least) but answering is a cop out because that raises the question "why is this definition compatible with another". To truly answer this we need to show i) the limit exists; call it v; ii)$db^x/dx = c_b b^x$ where c is a constant based on b; there is a value of b where $c_b $ =1. Call this value w. iii)call the anti derivative of 1/x, ln x; prove this is a logarithm based on a value z; show w=v=z=e. – fleablood Apr 03 '16 at 21:39

4 Answers4

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What you are correctly observing is:

$$\lim_{m \to \infty} \lim_{n \to \infty} (1+1/n)^m = \lim_{m \to \infty} 1 = 1.$$

To quickly see why something weird might happen, notice that

$$\lim_{n \to \infty} \lim_{m \to \infty} (1+1/n)^m = \lim_{n \to \infty} \infty = \infty.$$

In these two cases, we send $n,m$ to infinity separately and consecutively. In your limit, you send $n,m$ to infinity simultaneously and in fact at the same rate. Thus you are raising a number that is getting closer and closer to $1$ to a larger and larger power, and the two effects compete with each other to give a limit which is neither $1$ nor $\infty$.

Ian
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The reason is that $1 + 1/k$ is decreasing to $1$ while the exponent $k$ increases to $+\infty$, resulting in the indeterminate form $1^\infty$.

The expression $\lim_{k \to \infty} (1+1/k)^k$ is the definition of $e$.

Henricus V.
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Interpreting the question as why the limit is $>1$, the answer is that the sequence is strictly increasing. See The Limit Definition of e.

Martín-Blas Pérez Pinilla
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By the Taylor series expansion, as $x \to 0$, one obtains $$ \ln(1+x)=x+x\epsilon(x) $$ with $\displaystyle \epsilon(x) \to 0$, as $x \to 0$, giving as $n \to \infty$, $$ n\ln\left(1+\frac1n\right)=1+\epsilon\left(\frac1n\right) \to 1 $$ that is $$ \left(1+\frac1n\right)^n=e^{n\ln\left(1+\frac1n\right)} \to e^1=e. $$

Olivier Oloa
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