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I need to prove this approximation, but I am unable to conclude $$\log \left(1+\frac{1}{n}\right) \approx \frac{1}{n}$$

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closed as off-topic by user147263, user223391, user91500, Strants, user21820 Jul 15 '15 at 6:29

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    $\begingroup$ I'm confused. how does one prove an approximation ? How would I prove that $5$ is approximately equal to $17$ ? $\endgroup$ – mercio Jul 14 '15 at 15:26
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    $\begingroup$ What exactly does '$\approx$' mean? How close should LHS and RHS be? $\endgroup$ – BCLC Jul 14 '15 at 16:04
  • $\begingroup$ @mercio I suspect '$LHS \ \approx \ RHS$' is a heuristic for $|LHS - RHS| < \epsilon$ for some $\epsilon \in (0,1)$ $\endgroup$ – BCLC Jul 14 '15 at 16:04
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    $\begingroup$ 8 people think this question is well researched? $\endgroup$ – Alec Teal Jul 14 '15 at 20:31
  • $\begingroup$ possible duplicate of Showing y≈x for small x if y=log(x+1) $\endgroup$ – user147263 Jul 14 '15 at 21:12
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We have $\ln(x)=\lim_{m\to\infty}m(\sqrt[m]x-1)$.

By the generalized binomial theorem,

$$\lim_{m\to\infty}m\left(\sqrt[m]{1+\frac1n}-1\right)=\\ \lim_{m\to\infty}m\left(\frac1m\frac1n-\frac{m-1}{2m^2}\frac1{n^2}+\frac{(m-1)(2m-1)}{3!m^3}\frac1{n^3}-\frac{(m-1)(2m-1)(3m-1)}{4!m^4}\frac1{n^4}\cdots\right)\\ =\frac1n-\frac1{2n^2}+\frac1{3n^3}-\frac1{4n^4}\cdots$$

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  • $\begingroup$ The first two statements are much more complicated than the original question. $\endgroup$ – Teepeemm Jul 15 '15 at 14:04
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    $\begingroup$ @Teepeemm: glad to know. $\endgroup$ – Yves Daoust Jul 15 '15 at 14:05
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We have $$\log\left(1+\frac{1}{n}\right)=\log\left(n+1\right)-\log\left(n\right)=\int_{n}^{n+1}\frac{1}{t}dt $$ then by first mean value theorem for integration we have that exists some $c_{n}\in\left[n,n+1\right] $ such that $$=\frac{1}{c_{n}}\left(n+1-n\right)=\frac{1}{c_{n}}\approx\frac{1}{n}. $$

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    $\begingroup$ using integral test , wonderful proof $\endgroup$ – linux mint Jul 14 '15 at 6:22
  • $\begingroup$ In what sense do you mean ≈? How do you know OP means the same thing as you? $\endgroup$ – BCLC Jul 14 '15 at 17:26
  • $\begingroup$ @BCLC Normally $\approx$ means approximately. I don't know if you use it with another sense, but I always see it with this meaning. However I think it's quite obvious what I mean in this answer, and in fact OP doesn't ask more clarifications. $\endgroup$ – Marco Cantarini Jul 14 '15 at 17:44
  • $\begingroup$ Why the downvote? What's the problem? $\endgroup$ – Marco Cantarini Jul 15 '15 at 6:37
  • $\begingroup$ @MarcoCantarini When do you say two numbers are approximately equal? math.stackexchange.com/questions/1360393/… $\endgroup$ – BCLC Jul 16 '15 at 19:58
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The Taylor series for the function $\log(1+x)$ about the point $x=0$ is $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$ so when $|x|\ll 1$ we have $\log(1+x)\approx x$. See more at the Wikipedia article on Taylor's theorem.

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  • $\begingroup$ In what sense do you mean $\approx$? How do you know OP means the same thing as you? $\endgroup$ – BCLC Jul 14 '15 at 16:08
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You may apply the fact that, as $x \to a$, for any differentiable function around $a$, we have

$$ \frac{f(x)-f(a)}{x-a} \to f'(a). $$

Then take $f(x)=\log (1+x)$, with $f'(x)=\dfrac1{1+x}$, giving as $x \to 0$, $$ \frac{\log(1+x)-\log(1+0)}{x-0} \to \frac1{1+0}=1 $$ $$ \frac{\log(1+x)}x \to 1, $$ that is, as $n \to \infty$,

$$ \log\left(1+\frac1n\right) \sim \frac1n. $$

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  • $\begingroup$ I will review your proof. Thank you Olivier. =) $\endgroup$ – linux mint Jul 14 '15 at 6:24
  • $\begingroup$ @linuxmint You are welcome! $\endgroup$ – Olivier Oloa Jul 14 '15 at 6:26
  • $\begingroup$ Is "~" like the one here ? $\endgroup$ – BCLC Jul 14 '15 at 17:27
  • $\begingroup$ @BCLC I meant that the quotient tends to $1$ as the variable tends to $0$. $\endgroup$ – Olivier Oloa Jul 14 '15 at 17:29
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    $\begingroup$ @BCLC I meant the variable $x=\frac1n$, which tends to $0$. Thanks. $\endgroup$ – Olivier Oloa Jul 16 '15 at 20:10
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By definition,

$$e^{\ln(1+1/n)}=1+\frac1n.$$

Then, using the crude approximation* $e^x\approx1+x$,

$$1+\ln\left(1+\frac1n\right)\approx1+\frac1n.$$


*By the definition of $e$ and the binomial theorem,

$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n=\lim_{n\to\infty}1+x+\frac{n-1}{2n}x^2+\frac{(n-1)(n-2)}{3!n^2}x^3\cdots\approx1+x.$$

Using the next term will yield

$$1+\ln\left(1+\frac1n\right)+\frac12\ln^2\left(1+\frac1n\right)=1+\frac1n,$$ which can be solved for the logarithm.

$$\left(\ln\left(1+\frac1n\right)+1\right)^2=1+\frac2n.$$

A pretty contrived method.

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    $\begingroup$ @BCLC - How do you know we both see the same color red? $\endgroup$ – nbubis Jul 14 '15 at 16:29
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    $\begingroup$ @BCLC the notion $\approx$ is perfectly standard for anyone who has ever ventured outside the realm of pure mathematics. $\endgroup$ – nbubis Jul 14 '15 at 16:47
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    $\begingroup$ @nbubis Enlighten me, please? $\endgroup$ – BCLC Jul 14 '15 at 17:26
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    $\begingroup$ I wonder if I could use that crude approximation while building a bridge. It would save time. $\endgroup$ – PyRulez Jul 14 '15 at 17:41
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    $\begingroup$ The approximation $e^x\approx1+x$ clearly doesn't hold. One is linear the other is exponential. $\endgroup$ – kasperd Jul 14 '15 at 18:47
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Use the inequality $e^x\ge x+1$. Substitute in $x=\ln(1+\frac1n)$ and $x=-\ln(1+\frac1n)$ to obtain an upper and lower bound for $\ln(1+\frac1n)$.

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    $\begingroup$ Varying $n$, we get the infinite system of inequalities:$$1>\ln\frac21>\frac12>\ln\frac32>\frac13>\ln\frac43>\dotsb$$ $\endgroup$ – Akiva Weinberger Jul 14 '15 at 7:35

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