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I posted this question yesterday, and, despite getting answers, I am still confused how to solve it:

Use Taylor's theorem to prove that $\displaystyle\lim_{n \to \infty} n \ln\left(1+\frac{1}{n}\right)=1$

I know the Taylor expansion of ln(1+x) is $x-\frac{1}{2}x^2+ \frac{1}{3}x^3 ...$, but I don't see what this gives me.

Also, one respondent posted: ln(1+t)=t+o(t), and I don't understand how this follows by the Theorem...

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  • $\begingroup$ $\mathcal{O}(t)$ stands for the Big-O notation. You could search for it on Google. $\endgroup$ – user122283 Mar 4 '14 at 0:59
  • $\begingroup$ The power series for $\ln(1+x)$ converges absolutely for $|x|<1$. So the power series $1-\frac{1}{x}x+\frac{1}{3}x^{2}+\cdots$ does also, and this power series equals $\frac{1}{x}\ln(1+x)$ for $0 < |x| < 1$. A power series is continuous within its circle of convergence. So $\lim_{x\rightarrow 0}\frac{1}{x}\ln(1+x)$ is found by evaluating the series $1-\frac{1}{2}x+\frac{1}{3}x^{2}+\cdots$ at $x=0$. $\endgroup$ – Disintegrating By Parts Mar 4 '14 at 1:02
  • $\begingroup$ @NicholasR.Peterson It's the exact same question, by the same user. To quote the OP, "I posted this question yesterday,..." $\endgroup$ – user122283 Mar 4 '14 at 1:03
  • $\begingroup$ @SanathDevalapurkar That comment is automatically generated when you mark a question as a duplicate. $\endgroup$ – Nick Peterson Mar 4 '14 at 1:05
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    $\begingroup$ Please edit your previous post, or comment on the answers you've received, rather than asking a new copy. $\endgroup$ – user61527 Mar 4 '14 at 1:26
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Here $x = 1/n \implies \ln (1 + 1/n) = 1/n - \frac{1}{2}(\frac{1}{n})^2 + \frac{1}{3}(\frac{1}{n})^3 + \cdots \implies n\ln(1 + 1/n) = 1 - \frac{1}{2}(\frac{1}{n}) + \frac{1}{3}(\frac{1}{n})^2 + \cdots = 1 + O(\frac{1}{n})$. So $\lim\limits_{n\to \infty} n\ln (1 + 1/n) = 1$

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    $\begingroup$ Adding Latex would be helpful. $\endgroup$ – user122283 Mar 4 '14 at 1:03
  • $\begingroup$ Could you please explain how we get 1/n+0(1/n)? Sorry, I have never worked with this notation. Is there another way to write it? $\endgroup$ – kiwifruit Mar 4 '14 at 1:04
  • $\begingroup$ sure. I am working on mathjax now . $\endgroup$ – DeepSea Mar 4 '14 at 1:04
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    $\begingroup$ @kiwifruit $O(x)$ is just a way of saying that there are terms of higher power in $x$ which are usually included in the $\ldots$ left behind when expanding a series. $\endgroup$ – user122283 Mar 4 '14 at 1:07

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