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Find the value of \( \displaystyle \lim_{n \to \infty} \left (\displaystyle \prod_{r = 1}^{n} \left ( 1 + \frac{r}{n} \right)\right)^{\frac{1}{n}}\)

Note by Kushagraa Aggarwal 4 years, 1 month ago

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By properties of Riemann sums: \(\displaystyle\lim_{n \to \infty}\dfrac{1}{n}\sum_{r = 1}^{n}\ln \left(1 + \dfrac{r}{n}\right) = \int_{1}^{2}\ln(x)\,dx = \left[x\ln x - x\right]_{1}^{2} = 2\ln 2 - 1\).

Exponentiation yields \(\displaystyle\lim_{n \to \infty} \left[\prod_{r = 1}^{n}\left(1+\dfrac{r}{n}\right)\right]^{\tfrac{1}{n}} = e^{2 \ln 2 - 1} = \dfrac{4}{e}\).

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Isn't the product kinda discrete. I mean r and n can only take integer values, while the limit as n approaches infinity kinda assumes n can take continuous values. Is there something wrrong with this belief?

Hi,

He has mentioned there, Riemann sum

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TopNewestBy properties of Riemann sums: \(\displaystyle\lim_{n \to \infty}\dfrac{1}{n}\sum_{r = 1}^{n}\ln \left(1 + \dfrac{r}{n}\right) = \int_{1}^{2}\ln(x)\,dx = \left[x\ln x - x\right]_{1}^{2} = 2\ln 2 - 1\).

Exponentiation yields \(\displaystyle\lim_{n \to \infty} \left[\prod_{r = 1}^{n}\left(1+\dfrac{r}{n}\right)\right]^{\tfrac{1}{n}} = e^{2 \ln 2 - 1} = \dfrac{4}{e}\).

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Isn't the product kinda discrete. I mean r and n can only take integer values, while the limit as n approaches infinity kinda assumes n can take continuous values. Is there something wrrong with this belief?

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Hi,

He has mentioned there, Riemann sum

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