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How to find the following limit :

$$\lim_{n \to \infty}[(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})]^{\frac{1}{n}}$$ is equal to

(a) $\frac{4}{e}$

(b) $\frac{3}{e}$

(c) $\frac{1}{e}$

(d) $e$

Please suggest how to proceed in this problem thanks...

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marked as duplicate by Did, Umberto P., gerw, Davide Giraudo, M Turgeon Feb 21 '14 at 21:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

See this. – David Mitra Feb 21 '14 at 17:21
up vote 3 down vote accepted

$$\log\left(\lim_{n \to \infty}[(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})]^{\frac{1}{n}}\right) =\lim_{n \to \infty}\frac{\log(1+\frac{1}{n})+\log(1+\frac{2}{n})+\cdots+\log(1+\frac{n}{n})}{n} =\int_{1}^2 \log(1+x)dx= [x\log(x)-x]_{x=1}^{x=2}=2\log(2)-1$$

This yields the solution $e^{2\log(2)-1}=4/e$.

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Apply the $\log $ function and then use the Riemann sum.

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