ISI (3)

Calculus Level 3

lim ⁡ n → ∞ ( ∏ r = 1 n ( 1 + 2 r − 1 2 n ) ) 1 2 n = ? \lim_{n\to\infty} \left(\prod_{r=1}^n \left(1+\frac{2r-1}{2n}\right)\right)^{\frac 1{2n}}=?

1 e \frac{1}{\sqrt{e}} 2 e \frac{2}{\sqrt{e}} 4 e \frac{4}{\sqrt{e}} 2 e 2 \frac{2}{e^2} e e \frac{e}{\sqrt{e}}

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Chew-Seong Cheong
May 15, 2018

Relevant wiki: Riemann Sums

L = lim ⁡ n → ∞ ( ∏ r = 1 n ( 1 + 2 r − 1 2 n ) ) 1 2 n = exp ⁡ ( ln ⁡ ( lim ⁡ n → ∞ ( ∏ r = 1 n ( 1 + 2 r − 1 2 n ) ) 1 2 n ) ) where exp ⁡ ( x ) = e x = exp ⁡ ( lim ⁡ n → ∞ 1 2 n ∑ r = 1 n ln ⁡ ( 1 + 2 r − 1 2 n ) ) for n → ∞ = exp ⁡ ( lim ⁡ n → ∞ 1 2 n ∑ r = 1 n ln ⁡ ( 1 + r n ) ) By Riemann sums: = exp ⁡ ( 1 2 ∫ 0 1 ln ⁡ ( 1 + x ) d x ) lim ⁡ n → ∞ 1 n ∑ k = a b f ( k n ) = lim ⁡ n → ∞ ∫ a / n b / n f ( x ) d x = exp ⁡ ( 1 2 [ ( 1 + x ) ln ⁡ ( 1 + x ) − x ] 0 1 ) = exp ⁡ ( ln ⁡ 2 − 1 2 ) = 2 e \begin{aligned} L & = \lim_{n \to \infty} \left(\prod_{r=1}^n \left(1+\frac {2r-1}{2n}\right)\right)^{\frac 1{2n}} \\ & = \exp \left(\ln \left(\lim_{n \to \infty} \left(\prod_{r=1}^n \left(1+\frac {2r-1}{2n}\right)\right)^{\frac 1{2n}}\right)\right) & \small \color{#3D99F6} \text{where }\exp(x) = e^x \\ & = \exp \left(\lim_{n \to \infty} \frac 1{2n} \sum_{r=1}^n \ln \left(1+\frac {2r-1}{2n}\right) \right) & \small \color{#3D99F6} \text{for }n \to \infty \\ & = \exp \left(\lim_{n \to \infty} \frac 1{2n} \sum_{r=1}^n \ln \left(1+\frac rn \right) \right) & \small \color{#3D99F6} \text{By Riemann sums:} \\ & = \exp \left(\frac 12 \int_0^1 \ln (1+x) \ dx \right) & \small \color{#3D99F6} \lim_{n \to \infty} \frac 1n \sum_{k=a}^b f \left(\frac kn \right) = \lim_{n \to \infty} \int_{a/n}^{b/n} f(x) \ dx \\ & = \exp \left(\frac 12 \bigg[(1+x)\ln(1+x)-x\bigg]_0^1 \right) \\ & = \exp \left(\ln 2 - \frac 12 \right) \\ & = \boxed{\dfrac 2{\sqrt e}} \end{aligned}

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at last you lost your marathon streak.sorry to ask,how did that happen?

were you ill or lack of internet connection or you forgot?

Mohammad Khaza - 3 years ago

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Wow! You notice. I was holidaying in Europe with family. My phone wifi did not work well in the hotel in London. I thought I would lose it in country such as China or India. I will be travelling to China next week. I have kept the streak in Australia, China, Taiwan, Germany, France, Switzerland, Singapore, Thailand, Philippines even UK for one day other than my own country Malaysia. It is quite it will give me less stress when I travel overseas.

Chew-Seong Cheong - 3 years ago

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i hope, i will also travel the world one day and write books about the experience. stay fine.

Mohammad Khaza - 3 years ago

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