Mathematics

$$\displaystyle \int_1^e \log(x) dx= \displaystyle \lim_{n \rightarrow \infty} \displaystyle  \sum_{i=1}^n \log \left(1+i\dfrac{e-1}{n} \right)$$
State whether the above equation is True or False?


ANSWER

False


SOLUTION
Given : $$\displaystyle \int _{ 1 }^{ e }{ \ln { (x) } dx } = \lim_{ n\rightarrow \infty  }\sum _{ i=1 }^{ n }{ \log { \left[ 1+i\cfrac { (e-1) }{ n }  \right]  }  } $$
$$\displaystyle \int _{ 1 }^{ e }{ \ln { (x) } dx } =\left[ x\ln { x } -x \right] _{ 1 }^{ e }=e-e(D-1)=1\\ \displaystyle \lim_{ n\rightarrow \infty  } \sum _{ i=1 }^{ n }{ \log { \left[ 1+i\cfrac { (e-1) }{ n }  \right]  }  }$$
$$=\displaystyle \lim_{ n\rightarrow \infty  }\left[ \log { \left[ 1+i\cfrac { (e-1) }{ n }  \right]  } +\log { \left( 1+\cfrac { 2(e-1) }{ n }  \right)  } +\log { \left( 1+\cfrac { 3(e-1) }{ n }  \right)  } ........... \right] $$
$$= \displaystyle \lim_{ n\rightarrow \infty  } \left[ \cfrac { \log { \left[ 1+i\cfrac { (e-1) }{ n }  \right]  } \cfrac { (e-1) }{ n }  }{ \cfrac { (e-1) }{ n }  } +\cfrac { \log { \left( 1+\cfrac { 2(e-1) }{ n }  \right) \cfrac { 2(e-1) }{ n }  }  }{ \cfrac { 2(e-1) }{ n }  } +............ \right] $$
$$=\displaystyle \lim_{ n\rightarrow \infty  }\left( 1+2+3+4+........n \right) \left( \cfrac { (e-1) }{ n }  \right)$$
$$=\displaystyle \lim_{ n\rightarrow \infty  } \cfrac { n(n+1) }{ 2 } \cfrac { (e-1) }{ n } =\infty $$ 
Hence the statement is false.
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TRUE/FALSE Hard Published on 17th 09, 2020
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