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Can someone explain me why $\lim_{n \to \infty} n[\log (n+1)-\log(n)]=1$? Isn't that an indeterminate form? I mean, since $\lim_{n \to \infty} n = \infty$ and $\lim_{n \to \infty} [\log (n+1)-\log(n)] = 0$ then $\lim_{n \to \infty} a_n b_n=0 \cdot\infty$?

I'm just starting with sequences, and I have no idea what to do.

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    $\begingroup$ You can only say that $\lim a_n b_n = \lim a_n \cdot \lim b_n$, if both limits exist. If one is $\infty$, as in your example, the rule doesn't apply. $\endgroup$ – Viktor Glombik Dec 23 '18 at 22:12
  • $\begingroup$ You may think of $\log n$ as $\int_{1}^{\infty}\frac1x dx$. $\endgroup$ – MR_BD Dec 24 '18 at 15:09
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As an alternative without L'Hopital, but rather using the very well known limit: $$ \begin{align} \lim_{n\to\infty}\left(n\cdot\left(\ln(n+1) - \ln n\right)\right) &= \lim_{n\to\infty}\left(n\cdot\ln\left(\frac{n+1}{n}\right)\right) \\ &= \lim_{n\to\infty} \ln\left(\frac{n+1}{n}\right)^n\\ &= \ln \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\\ &= \ln e \\ &= 1 \end{align} $$

I'm assuming you used $\log x$ for $\log_ex$ or simply $\ln x$.

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If you already know the rule of L'HOSPITAL, the calculation could go as follows \begin{align*} L := \lim_{n \to \infty} n \cdot \log \left( \frac{n + 1}{n} \right) = \lim_{n \to \infty} \frac{\log \left( \frac{n + 1}{n} \right)}{\frac{1}{n}} \end{align*} (here I am only using logarithm and fraction properties)

Now this in an indeterminate form ''$\frac{0}{0}$'', so we can apply the above mentioned rule by differentiating the numerator and the denominator: \begin{align*} L = \lim_{n \to \infty} \frac{\frac{n}{n + 1} \left( \frac{1}{n} -\frac{n + 1}{n^2} \right)}{- \frac{1}{n^2}} = \lim_{n \to \infty} \frac{n}{n + 1} = 1. \end{align*} This follows because \begin{equation*} \frac{d}{dx} \frac{1}{x} = \frac{d}{dx} x^{-1} = - x^{-2} \end{equation*} and \begin{equation*} \frac{d}{dx} \log\left(\frac{x + 1}{x} \right) = \frac{d}{dx} \log(x + 1) - \frac{d}{dx} log(x) = \frac{1}{x + 1} - \frac{1}{x} = \frac{x}{x + 1} \left( \frac{1}{x} -\frac{x + 1}{x^2} \right) \end{equation*}

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It's because it can rewritten as $$n\log\frac{n+1}n=n\log\Bigl(1+\frac1n\Bigr)=\frac{\log\Bigl(1+\dfrac1n\Bigr)}{\dfrac1n}$$ Set $u=\dfrac1n$. This expression becomes $\dfrac{\log(1+u)}u$, and it is standard from high school that the limit of this quotient as $u\to 0$ is$\;\bigl(\log(1+u)\bigr)'_{u=0},\:$ i.e. $\:1$.

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You could even go beyond the limit. Write $$a_n= n[\log (n+1)-\log(n)]=n \log\left(\frac{n+1}n \right)=n \log\left(1+\frac{1}n \right)$$ Now, remembering that for small $x$ $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right)$$ make $x=\frac 1 n$ to get $$a_n=n\left(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right)\right)=1-\frac{1}{2 n}+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$

Just try it for $n=10$. You will get $$a_{10}=10 \,\log \left(\frac{11}{10}\right)\approx 0.953102$$ while the above expansion would give $$a_{10}\simeq \frac{143}{150}\approx 0.953333$$

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  • $\begingroup$ This time faster than @gimusi :) $\endgroup$ – roman Dec 24 '18 at 11:38
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    $\begingroup$ @roman. For once !! By the way, Merry Xmas $\endgroup$ – Claude Leibovici Dec 24 '18 at 11:40

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