# $\lim_{x \to +\infty} \frac{\ln(x)}{x}$ using Taylor series

How can I find the limit $$\lim_{x \to +\infty} \frac{\ln(x)}{x}$$ using Taylor series?

I know that $$\ln(1+x) = 1- x + x^2 + (-1)^n x^n + O(x^{n+1})$$, so what I did was:

$$\ln(x) = \ln(1+(x-1)) = 1-(x-1) + O((x-1)^2)$$

$$\displaystyle \lim_{x \to +\infty} \frac{1-(x-1) + O((x-1)^2)}{x}$$

But not sure how to proceed from here...

• Bad idea to taylor expand the logarithm around $\infty$... The power series for the logarithm only converges for $|x|<1$ Commented May 17, 2014 at 15:50
• Your taylor series only converges for $|x| < 1$. This means, you cannot use it to calculate the limit as $x \rightarrow \infty$. Commented May 17, 2014 at 15:51

As your Taylor series converges only for $$x<1$$, try using $$1/x$$ instead of $$x$$: $$\lim_{x \to 0} \frac{\ln(\frac 1 x)}{\frac1 x} = \lim_{x \to 0} (-x\ln(x))$$

Okaaay... I am a bit at loss but, I guess, now if we want that badly to use Taylor series — we should just go for it. First, let's build the series for the whole function. \begin{align} (x \ln x)' &= \ln x + 1 \\ (x \ln x)'' &= (\ln x + 1)' = \frac 1 x \\ (x \ln x)''' &= \left(\frac 1 x\right)' = - \frac 1 {x^2} \\ (x \ln x)'''' &= \left(-\frac 1 {x^2}\right)' = 2 \frac 1 {x^3} \\ (x \ln x)''''' &= - 3! \frac 1 {x^4} \end{align}

and so on, with the $$n$$th derivative equal to $$(-1)^n(n-2)!\frac 1 {x^{n-1}}$$ (works for the derivatives starting with the third one, of course)

This gives us Taylor series for $$(1+x)\ln x$$: $$0 + x + \frac {x^2} {2!}-\frac {x^3} {3!}+\frac {2x^4} {4!}-\frac {3!x^5} {5!}+\frac {4!x^6} {6!} - \frac {5!x^7} {7!}+\ldots$$ So the later terms are $$(-1)^nx^n\frac 1 {n(n-1)}$$. The expression above does have a finite sum for $$x=-1$$, so the series for $$x\to-1$$ has a limit equal to the sum: $$\lim _{y\to0}{y\ln y} = -1 + \frac 1 2 + \frac 1 6 + \frac 1 {12} + \frac 1 {20}+\ldots = -1 + \sum _{n=2}^{\infty}{\frac 1 {n(n-1)}}$$

Mathematical induction can easily be used to prove that a partial sum of the last series is $$S_m = \frac {m-1} m$$. This means that the infinite sum $$\sum _{n=2}^{\infty}{\frac 1 {n(n-1)}} = \lim_{m\to\infty}{\frac {m-1} m}=1$$

So $$\lim _{x\rightarrow+\infty}{\frac {\ln x} x}=-\lim _{y\to0}{y\ln y} = -\left(-1 + \sum _{n=2}^{\infty}{\frac 1 {n(n-1)}}\right)= 1 -1 = 0$$ Which was obvious from the beginning, but it is nice that one can prove that in such a complicated way.

• Your comment " it is nice that one can prove that in such a complicated way" is quite funny: +1. Commented May 31, 2014 at 20:13

Let $$x=\frac1{1-u}$$. Note that as $$u\to1^-$$, we have $$x\to\infty$$. \begin{align} \lim_{x\to\infty}\frac{\log(x)}x &=\lim_{u\to1^-}(1-u)\log\left(\frac1{1-u}\right)\\ &=\lim_{u\to1^-}(1-u)\sum_{k=1}^\infty\frac{u^k}k\\ &=\lim_{u\to1^-}\left[u-\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)u^{k+1}\right]\\ &=1-\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)\\[6pt] &=0 \end{align}