We know that limit $\left(1+\dfrac{x}{n}\right)^n$ converges to $e^x$ but how can we prove that limit $\left(1+\dfrac{x}{n}+o(\frac{1}{n})\right)^n$ converges to $e^x$.
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what does $o(1/n)$ represent? – Varun Iyer Jul 03 '14 at 16:49
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O(1/n) represents the order of 1/n terms . – Bob Jul 03 '14 at 16:54
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Don't make a mess with $o$ and $O.$ – mfl Jul 03 '14 at 17:03
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Actually i also have some confusion with these notations o and O and their meaning. Can you suggest some link where should i get these things. – Bob Jul 03 '14 at 17:31
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http://math.stackexchange.com/questions/374747/if-z-n-to-z-then-1z-n-nn-to-ez – Jul 03 '14 at 18:44
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1Please, "limits" do not "converge", sequences converge (or not) and (when they converge) limits are equal to something. – Did Jul 03 '14 at 19:28
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Suppose that $f(n)=o(1/n)$.
Choose any $\epsilon\gt0$. There is an $N$ so that if $n\ge N$, then $\left|f(n)\right|\le\epsilon/n$. Therefore, $$ e^{x-\epsilon}=\lim_{n\to\infty}\left(1+\frac{x-\epsilon}n\right)^n\le\lim_{n\to\infty}\left(1+\frac xn+f(n)\right)^n\le\lim_{n\to\infty}\left(1+\frac{x+\epsilon}n\right)^n=e^{x+\epsilon} $$ Since $\epsilon\gt0$ was arbitrary, we have $$ \lim_{n\to\infty}\left(1+\frac xn+f(n)\right)^n=e^x $$
As gniourf_gniourf mentions, we also need to choose $N$ big enough so that for $n\ge N$, $$ 1+\frac{x-\epsilon}{n}\ge0 $$ since $x\le y\implies x^n\le y^n$ when $x,y\ge0$.
robjohn
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How do you justify $$\lim_{n\to+\infty}\left(1+\frac{x-\varepsilon}n\right)^n\leq\lim_{n\to+\infty}\left(1+\frac xn+f(n)\right)^n?$$ you must also mention that you take $n$ sufficiently large for $\dfrac{x-\varepsilon}n$ to be greater than $-1$ so that you're in the domain where $t\mapsto t^n$ is increasing. – gniourf_gniourf Jul 03 '14 at 17:33
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For $\alpha_n \to 0$ as $n\to \infty$ we have
$$\lim_{n\to \infty}\left(1+\frac{x}{n}+\alpha_n\right)^n=e^{\lim_{n\to \infty} (x+n\alpha_n)}=e^x,$$ where in the last equality we have used that $\alpha_n=o(1/n),$ that is, $\lim_n n\alpha_n=0.$
Edit
- Justification of the identity used:
Assume $\alpha_n\to 0, \beta_n\to \infty$ and that $\lim_n\to \infty$ exits. Since $e=\lim_n (1+\alpha_n)^{1/\alpha_n}$ we have that
$$\displaystyle \lim_n (1+\alpha_n)^{\beta_n}=\lim_n (1+\alpha_n)^{ \frac{\alpha_n\beta_n}{\alpha_n}}=\left(\lim_n (1+\alpha_n)^{1/\alpha_n}\right)^{\lim_n \alpha_n\beta_n}=e^{\lim_n \alpha_n\beta_n}.$$
- $o()$-notation:
$\alpha_n=o(1/n)$ means, by definition, that $\lim_n \frac{\alpha_n}{1/n}=0.$ But $\lim_n \frac{\alpha_n}{1/n}=\lim_n n\alpha_n$ and so it exists and is $0.$
mfl
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And i want a proof for this thing using $e^x$ and the the limit of the above term tending to zero. – Bob Jul 03 '14 at 17:07
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How do you justify that $$\lim_{n\to+\infty}\left(1+\frac xn+\alpha_n\right)^n=\exp\left(\lim_{n\to+\infty}(x+n\alpha_n)\right)?$$ – gniourf_gniourf Jul 03 '14 at 17:27
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@Did Yes, both limits exist and they are zero. http://mathworld.wolfram.com/Little-ONotation.html – mfl Jul 03 '14 at 22:12
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The problem is in the way you present things, not in the result or in definitions. – Did Jul 04 '14 at 05:21
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Fix $x\in\mathbb{R}$.
Let $n\in\mathbb{N}^*$. Then: $$\left(1+\frac xn+o\left(\frac1n\right)\right)^n=\exp\left(n\ln\left(1+\frac xn+o\left(\frac1n\right)\right)\right).$$ Now, since $$\dfrac xn+o\left(\dfrac1n\right)\underset{n\to+\infty}\longrightarrow0$$ and $$o\left(\dfrac xn+o\left(\dfrac1n\right)\right)\underset{n\to+\infty}=o\left(\dfrac 1n\right),$$ we have $$\ln\left(1+\frac xn+o\left(\frac1n\right)\right)\underset{n\to+\infty}=\frac xn+o\left(\frac1n\right),$$ hence $$n\ln\left(1+\frac xn+o\left(\frac1n\right)\right)\underset{n\to+\infty}=x+o(1),$$ hence $$\lim_{n\to+\infty}n\ln\left(1+\frac xn+o\left(\frac1n\right)\right)=x.$$ Now you can conclude by composing by the exponential function (which is continuous).
gniourf_gniourf
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