# Proof that $\lim_{n \to \infty} n[\log (n+1)-\log(n)]=1$

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.

• 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. – Viktor Glombik Dec 23 '18 at 22:12
• You may think of $\log n$ as $\int_{1}^{\infty}\frac1x dx$. – MR_BD Dec 24 '18 at 15:09

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 $$\ln x$$.

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 $$\begin{equation*} \frac{d}{dx} \log\left(\frac{x + 1}{x} \right) = \frac{d}{dx} \log(x + 1) - \frac{d}{dx} log(x) = \frac{1}{x + 1} - \frac{1}{x} = \frac{x}{x + 1} \left( \frac{1}{x} -\frac{x + 1}{x^2} \right) \end{equation*}$$

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. $$\:1$$.

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 $$a_{10}\simeq \frac{143}{150}\approx 0.953333$$

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