MHB How Can I Find the Limit of the Natural Log Function in the Form of a Question?

[anemone]

Find the

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

 

[Sudharaka]

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]

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]

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]

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

 

[CaptainBlack]

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

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

CB

 

[chisigma]

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]

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