Limit.

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

#Limits #MathProblem #Math

Easy Math Editor

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Jimmy Kariznov - 7 years, 9 months ago

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?

Sebastian Garrido - 7 years, 9 months ago

Hi,

He has mentioned there, Riemann sum

gopinath no - 7 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$