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Evaluate $$\lim_{n\to\infty} \left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}\right)...........\left(1+\frac{n}{n}\right)\right]^\frac{1}{n}$$ I have tried to solve this problem by taking logarithm but it works for finding the limit of functions but not in the case of sequences the answer to the problem is 4/e

Mathematics 13
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    This has been asked many times before. Please perform a search on the site: http://math.stackexchange.com/questions/1071053/to-evaluate-limit-of-sequence-left-left-1-frac1n-right-left-1-frac. – StubbornAtom Nov 04 '16 at 12:22

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Let $L$ be your limit, and take the logarithm. You get $$ \ln L = \lim_{n \to \infty}\frac1n\sum_{i = 1}^n\ln\left(1+\frac in\right) $$ where the right-hand side is an upper Darboux sum converging to the integral $$ \int_0^1\ln(1+x)dx =\left[-x+x\ln(1+x) + \ln(1+x)\right]_{x = 0}^1 = \ln 4 - 1 $$ So we have $\ln L = \ln 4 -1$, which means $L = e^{\ln 4 - 1} = \frac4e$

Arthur
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We can write this as ((2n!)/(n!)^2)^(1/n) which can be then simplified by using simple limit formula to get 4/e when n tends to infinity.