Limit of natural log

anemone

MHB POTW Director
Staff member
Find the

$\displaystyle\lim_{n\to\infty}\frac{2ln(2)+3ln(3)+......+nln(n)}{n^2ln(n)}$

Sudharaka

Well-known member
MHB Math Helper
Find the

$\displaystyle\lim_{n\to\infty}\frac{2ln(2)+3ln(3)+......+nln(n)}{n^2ln(n)}$
Hi anemone,

Use the logarithm rules, $$\ln (ab)=\ln (a)+\ln (b)\mbox{ and }\ln (a^n)=n\ln (a)$$ on the numerator and try to simplify it.

chisigma

Well-known member
Find the

$\displaystyle\lim_{n\to\infty}\frac{2ln(2)+3ln(3)+......+nln(n)}{n^2ln(n)}$
A comfortable solution is given from the Stolz-Cesaro Theorem that extablishes, given two sequences $a_{n}$ and $b_{n}$, that...

$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}} = \lim_{n \rightarrow \infty} \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}$ (1)

... if the limits in (1) exist. In that case $\displaystyle a_{n}=\sum_{k=1}^{n} k\ \ln k$ and $\displaystyle b_{n}=n^{2}\ \ln n$ so that is...

$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}}= \lim_{n \rightarrow \infty} \frac{(n+1)\ \ln (n+1)}{(n+1)^{2}\ \ln (n+1) - n^{2}\ \ln n}= \lim_{n \rightarrow \infty} \frac{n+1}{2n+1}= \frac{1}{2}$ (2)

Kind regards

$\chi$ $\sigma$

CaptainBlack

Well-known member
Find the

$\displaystyle\lim_{n\to\infty}\frac{2ln(2)+3ln(3)+......+nln(n)}{n^2ln(n)}$
One way of doing this is to relate the sum: $$\sum_{k=2}^n k\ln(k)$$ to the integral $$\int_{x=2}^n x\ln(x)\;dx$$ where we note that the integrand in the latter is an increasing function.

CB

sbhatnagar

Active member
\begin{align*}\lim_{n \to \infty}\sum_{r=1}^{n}\frac{r \ln(r)}{n^2 \ln(n)} &= \lim_{n \to \infty} \frac{1}{\ln(n)} \left[ \sum_{r=1}^{n}\frac{r}{n^2} \left\{ \ln{\left( \frac{r}{n}\right)+\ln(n)}\right\}\right] \\ &= \lim_{n \to \infty} \frac{1}{\ln(n)} \left[ \sum_{r=1}^{n}\frac{r}{n^2}\ln{\left( \frac{r}{n}\right)}+ \sum_{r=1}^{n}\frac{r\ln(n)}{n^2}\right] \\ &= \lim_{n \to \infty}\frac{1}{\ln(n)} \left[ \int_{0}^{1}x \cdot \ln(x) \, dx + \frac{\ln(n)}{n^2}\sum_{r=1}^{n}r\right] \\ &=\lim_{n \to \infty}\frac{1}{\ln(n)} \left[ -\frac{1}{4} + \frac{n(n+1)\ln(n)}{2n^2}\right] \\ &=\lim_{n \to \infty} \left[ -\frac{1}{4\ln(n)}+\frac{1}{2}+\frac{1}{2n}\right] \\ &=0+\frac{1}{2}+0 \\ &=\frac{1}{2} \end{align*}

CaptainBlack

Well-known member
\begin{align*}\lim_{n \to \infty}\sum_{r=1}^{n}\frac{r \ln(r)}{n^2 \ln(n)} &= \lim_{n \to \infty} \frac{1}{\ln(n)} \left[ \sum_{r=1}^{n}\frac{r}{n^2} \left\{ \ln{\left( \frac{r}{n}\right)+\ln(n)}\right\}\right] \\ &= \lim_{n \to \infty} \frac{1}{\ln(n)} \left[ \sum_{r=1}^{n}\frac{r}{n^2}\ln{\left( \frac{r}{n}\right)}+ \sum_{r=1}^{n}\frac{r\ln(n)}{n^2}\right] \\ &= \lim_{n \to \infty}\frac{1}{\ln(n)} \left[ \int_{0}^{1}x \cdot \ln(x) \, dx + \frac{\ln(n)}{n^2}\sum_{r=1}^{n}r\right] \\ &=\lim_{n \to \infty}\frac{1}{\ln(n)} \left[ -\frac{1}{4} + \frac{n(n+1)\ln(n)}{2n^2}\right] \\ &=\lim_{n \to \infty} \left[ -\frac{1}{4\ln(n)}+\frac{1}{2}+\frac{1}{2n}\right] \\ &=0+\frac{1}{2}+0 \\ &=\frac{1}{2} \end{align*}
The replacement of the Riemann sum by the integral inside the limit when going from line 2 to 3 needs more justification.

CB

chisigma

Well-known member
The replacement of the Riemann sum by the integral inside the limit when going from line 2 to 3 needs more justification.

CB
In $(0,1]$ the function $f(x)= x\ \ln x$ is bounded [more precisely $0 \le |f(x)| \le \frac{1}{e}$...] and continuous so that is Riemann-integrable...

Kind regards

$\chi$ $\sigma$

CaptainBlack

Well-known member
In $(0,1]$ the function $f(x)= x\ \ln x$ is bounded [more precisely $0 \le |f(x)| \le \frac{1}{e}$...] and continuous so that is Riemann-integrable...

Kind regards

$\chi$ $\sigma$
1. I know the limit/integral exists

2. I know that the method can be made to work

2. You cannot replace a finite sum by the limit, inside the limit that way, it needs further justification, not much but some: The LaTeX is now unreadable, but used something like:

$\lim_{n \to \infty} \left( a_n b_n \right) =\left(\lim_{n \to \infty}a_n\right)\left(\lim_{n \to \infty}b_n\right)$

CB

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