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