A calculus problem by Jeremy marquardt

The limit can be written as $L=\lim \limits_{n\to \infty }\frac{ln\left( 1\times 2\times 3\times \cdots \times n\right) -ln\left( n^{n}\right) }{n} =\lim \limits_{n\to \infty }\frac{ln\left( \frac{n! }{n^n} \right) }{n}$ Now using Stirling formula to get the limit $L=\lim \limits_{n\to \infty }\frac{ln\left( \frac{\sqrt{2\pi n} n^{n}}{n^{n}e^{n}} \right) }{n} =L_{1}+L_{2}=\lim \limits_{n\to \infty }\frac{ln\left( \sqrt{2\pi n} \right) }{n} -\lim \limits_{n\to \infty }\frac{ln\left( e^{n}\right) }{n}$ For the first limit we will use L'Hopital Rule And the second will equal to $-1$ Finally $L_{1} =\lim \limits_{n\to \infty }\frac{ln\left( \sqrt{2\pi n} \right) }{n} =\lim \limits_{n\to \infty }\frac{1}{2n} =0$ And $L_{2}=\lim \limits_{n\to \infty }\frac{ln\left( e^{n}\right) }{n} =1$ Therefore $L=0-1=\boxed{-1}$

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$\lim_{n\rightarrow \infty } \frac{ln(1)+ln(2)+ln(3)+...+ln(n)}{n}-ln(n)$

which leads to

$\lim_{n\rightarrow \infty } \frac{ln(1)+ln(2)+ln(3)+...+ln(n)-n*ln(n)}{n}$

Since there are n terms of the sum, and n -ln(n)s, there is exactly one -ln(n) for every part of the series, meaning

$\lim_{n\rightarrow \infty } \frac{1}{n}*[ln(\frac{1}{n})+ln(\frac{2}{n})+ln(\frac{3}{n})+...+ln(\frac{n}{n})]$ (brackets used for clarity)

$\lim_{n\rightarrow \infty } \frac{1}{n}*\sum_{x=1}^{n} ln(\frac{x}{n})$

Can be viewed as a right Riemann sum that becomes more and more accurate to ln(x) from 0 to 1

$\int_{0}^{1} ln(x) dx$

which evaluates to

$-1$

Since the function $\ln x$ is monotonic increasing on $(0,\infty)$ , we see that $L_n \; = \; \tfrac{1}{n}\sum_{r=1}^n \ln\big(\tfrac{r}{n}\big) \; \ge \; \sum_{r=1}^n \int_{\frac{r-1}{n}}^{\frac{r}{n}} \ln x\,dx \; = \; \int_0^1 \ln x\,dx \; = \; -1$ Also $L_n \; \le \; \sum_{r=1}^n \int_{\frac{r}{n}}^{\frac{r+1}{n}} \ln x\,dx \; = \; \int_{\frac{1}{n}}^{1 + \frac{1}{n}} \ln x\,dx$ and hence $-1 \le L_n \le \Big[x \ln x - x\Big]_{\frac{1}{n}}^{1+\frac{1}{n}} \; = \; \big(1 + \tfrac{1}{n}\big)\ln\big(1 + \tfrac{1}{n}\big) - 1 - \tfrac{1}{n} + \frac{\ln n}{n} + \tfrac{1}{n} \; = \; \ln\big(1 + \tfrac{1}{n}\big) + \frac{\ln(n+1)}{n} - 1$ so it is clear that $\lim_{n \to \infty} L_n \;= \; -1$

Let the value of the given limit be L. $\Rightarrow L=\lim _{ n\rightarrow \infty }{ \frac { \ln { 1 } +\ln { 2 } +\ln { 3 } +\cdots +\ln { n } }{ n } } -\ln { n }$ $\Rightarrow L=\lim _{ n\rightarrow \infty } \frac { 1 }{ n } \left( \ln { \frac { 1 }{ n } } +\ln { \frac { 2 }{ n } } +\ln { \frac { 3 }{ n } } \cdots+ \ln { \frac { n }{ n } } \right)$ $\Rightarrow L=\lim _{ n\rightarrow \infty } \sum _{ r=1 }^{ n }{ \frac { 1 }{ n } \ln { \frac { r }{ n } } }$ Taking $\frac{r}{n}=x$ , then $\frac{1}{n}=dx$ , we can convert it into a definite integral with upper limit $\rightarrow \frac{ n }{ n }=1$ and lower limit $\rightarrow \frac{ 1 }{ n }=0$ $\Rightarrow L=\int _{ 0 }^{ 1 }{ \ln { x } .dx } =[x\ln{x}-x]_{0}^{1}=\left( -1 \right) -0=\boxed { -1 }$

Used stolz cesaro theorem..