Friday, January 8, 2016

sequences and series - Limits Problem : $lim_{n to infty}[(1+frac{1}{n})(1+frac{2}{n})cdots(1+frac{n}{n})]^{frac{1}{n}}$ is equal to..



Problem:



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...


Answer



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


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



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