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- Feb 14, 2012
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Find the
$\displaystyle\lim_{n\to\infty}\frac{2ln(2)+3ln(3)+......+nln(n)}{n^2ln(n)}$
$\displaystyle\lim_{n\to\infty}\frac{2ln(2)+3ln(3)+......+nln(n)}{n^2ln(n)}$
Hi anemone,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...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.Find the
$\displaystyle\lim_{n\to\infty}\frac{2ln(2)+3ln(3)+......+nln(n)}{n^2ln(n)}$
The replacement of the Riemann sum by the integral inside the limit when going from line 2 to 3 needs more justification.$$\begin{align*}\lim_{n \to \infty}\sum_{r=1}^{n}\frac{r \ln(r)}{n^2 \ln(n)} &= \lim_{n \to \infty} \frac{1}{\ln(n)} \left[ \sum_{r=1}^{n}\frac{r}{n^2} \left\{ \ln{\left( \frac{r}{n}\right)+\ln(n)}\right\}\right] \\ &= \lim_{n \to \infty} \frac{1}{\ln(n)} \left[ \sum_{r=1}^{n}\frac{r}{n^2}\ln{\left( \frac{r}{n}\right)}+ \sum_{r=1}^{n}\frac{r\ln(n)}{n^2}\right] \\ &=
\lim_{n \to \infty}\frac{1}{\ln(n)} \left[ \int_{0}^{1}x \cdot \ln(x) \, dx + \frac{\ln(n)}{n^2}\sum_{r=1}^{n}r\right] \\
&=\lim_{n \to \infty}\frac{1}{\ln(n)} \left[ -\frac{1}{4} + \frac{n(n+1)\ln(n)}{2n^2}\right] \\
&=\lim_{n \to \infty} \left[ -\frac{1}{4\ln(n)}+\frac{1}{2}+\frac{1}{2n}\right] \\
&=0+\frac{1}{2}+0 \\
&=\frac{1}{2}
\end{align*}$$
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...The replacement of the Riemann sum by the integral inside the limit when going from line 2 to 3 needs more justification.
CB
1. I know the limit/integral existsIn $(0,1]$ 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
$\chi$ $\sigma$