Question

${\int }_1^e\log \left(x\right)dx=\underset{n\to \infty }{lim}\sum _{i=1}^n\log (1+i\frac{e-1}{n})$
State whether the above equation is True or False?

• A. True
• B. False

Answer 1

The correct option is B False

Given : ${\int }_1^e\ln \left(x\right)dx=\underset{n\to \infty }{lim}\sum _{i=1}^n\log [1+i\frac{(e-1)}{n}]$

${\int }_1^e\ln \left(x\right)dx={[x\ln x-x]}_1^e=e-e(D-1)=1\underset{n\to \infty }{lim}\sum _{i=1}^n\log [1+i\frac{(e-1)}{n}]$

$=\underset{n\to \infty }{lim}[\log [1+i\frac{(e-1)}{n}]+\log (1+\frac{2(e-1)}{n})+\log (1+\frac{3(e-1)}{n})...........]$

$=\underset{n\to \infty }{lim}[\frac{\log [1+i\frac{(e-1)}{n}]\frac{(e-1)}{n}}{\frac{(e-1)}{n}}+\frac{\log (1+\frac{2(e-1)}{n})\frac{2(e-1)}{n}}{\frac{2(e-1)}{n}}+............]$

$=\underset{n\to \infty }{lim}(1+2+3+4+........n)\left(\frac{(e-1)}{n}\right)$

$=\underset{n\to \infty }{lim}\frac{n(n+1)}{2}\frac{(e-1)}{n}=\infty$

Hence the statement is false.