\lim_{n\rightarrow \infty }\sum_{r=1}^{n}\frac{1}{n}e^{r/n}\;\; \; \; is

  • Option 1)

    1-e

  • Option 2)

    e-1

  • Option 3)

    e

  • Option 4)

    e+1

 

Answers (1)

As we learnt in 

Walli's Method -

Definite integral by first principle

\int_{a}^{b}f(x)dx= \left ( b-a \right )\lim_{n \to \infty }\frac{1}{n}\left [ f(a) +f(a+h)+f(a+2h)....\right ]

where

h=\frac{b-a}{n}

- wherein

 

 

\lim_{n\rightarrow \infty} \sum_{r=1}^{n}\frac{1}{n}e^{r/n}

\Rightarrow \int_{0}^{1}e^n dn

\Rightarrow \left [ e^n \right ]_{0}^{1}

\Rightarrow e-1


Option 1)

1-e

This is incorrect option

Option 2)

e-1

This is correct option

Option 3)

e

This is incorrect option

Option 4)

e+1

This is incorrect option

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