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Evaluate : $\lim_{n\to \infty}\{(1+\frac{1}{2n})(1+\frac{3}{2n})(1+\frac{5}{2n})\cdots (1+\frac{2n-1}{2n})\}^{\frac{1}{2n}}$

My attempt : Since, [$\displaystyle \lim\limits_{n\to \infty} a_n^{1/n} = \lim\limits_{n\to \infty} \frac{a_{n+1}}{a_n}$] $$Here \ \ a_n = \prod\limits_{k=1}^{n} \left(1+\frac{2k-1}{2n}\right) $$ I got $$\frac{ a_{n+1}}{a_n}=\frac{(4n+1)(4n+3)}{(2n+1)(2n+2)\times (1+\frac{1}{n})^n}$$ $$\ Hence, \ \lim\limits _{n\to \infty} {\frac {a_{n+1}}{a_n}}=\frac{4}{e} $$

My question is can I still use Cauchy's second theorem if the power of the function is $\frac{1}{2n}$? Caused I've solved it same as the power $\frac{1}{n}$. Please help me for this. Thank you in advance.

Chris
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2 Answers2

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Hint

Let $$x_n=\prod_{k=0}^{n-1}\left(1+\frac{2k+1}{2n}\right)^{\frac{1}{2n}}\quad \text{and}\quad y_n=\prod_{k=0}^{n-1}\left(1+\frac{2k}{2n}\right)^{\frac{1}{2n}}.$$

let $$k_n=\prod_{k=0}^{2n-1}\left(1+\frac{k}{n}\right)^\frac{1}{2n}.$$ So, $$x_n=\frac{k_n}{y_n},$$ and you can compute $$\lim_{n\to \infty }\ln(y_n)\quad \text{and}\quad \lim_{n\to \infty }\ln(k_n),$$ using Riemann sum.

Surb
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Another way $$:L\lim_{n\to \infty}\{(1+\frac{1}{2n})(1+\frac{3}{2n})(1+\frac{5}{2n})\cdots (1+\frac{2n-1}{2n})\}^{\frac{1}{2n}}$$ $$\implies \ln L=lim_{n\to \infty}\frac{1}{2n} \sum_{k=1}^{n} \ln \left(1+\frac{2k-1}{2n}\right)=\frac{1}{2} \int_{0}^{1} \ln (1+x) dx$$ Here, we use $k/n=x, 1/n \to dx$ when $n$ is very large. $$\implies \ln L=\frac{1}{2} [(1+x)\ln(1+x)-(1+x)]_{0}^{1}=\ln 2-1/2$$ $$L=2 e^{-1/2}.$$

Z Ahmed
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  • In the step $$\ln L=\lim_{n\to \infty}\frac{1}{2n} \sum_{k=1}^{n} \ln \left(1+\frac{2k-1}{2n}\right)=\frac{1}{2} \int_{0}^{1} \ln (1+x) dx$$ Have you taken $$\frac{2k-1}{2n}=\frac{k}{n}$$ and neglected $\frac{1}{n}$? – DatBoi Aug 19 '20 at 13:02
  • Yes, as $n$ is very large. – Z Ahmed Aug 19 '20 at 13:49
  • In that case, we are adding 0's infinite times. The sum may or may not be tending to 0. Am I missing something? – DatBoi Aug 19 '20 at 13:51
  • Yes, you are right $\ln L$ actually takes the indeterminate form $0\times \infty$, so one requires a special method of converting the limit of sum to an integral when $n\to \infty$. – Z Ahmed Aug 19 '20 at 14:30
  • What is that method may I ask? – DatBoi Aug 19 '20 at 14:35
  • $\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n f(k/n)= \int_{0}^{1} f(x) dx$ $k/n=x$ and $1/n \to dx$, sum is changed to integral.. – Z Ahmed Aug 19 '20 at 14:39
  • Please don't get me wrong. I know that $$\int_0^1{f(x)dx}=\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^n{f(\frac{k}{n})}$$ In this case we have $$\lim_{n\to \infty}\frac{1}{2n} \sum_{k=1}^{n} \ln \left(1+\frac{2k-1}{2n}\right)$$ But $\lim_{n\to \infty}\frac{1}{2}\frac{1}{n} \sum_{k=1}^{n} \ln \left(1+\frac{k}{n}\right=\frac{1}{2}\int_0^1{ln(1+x)dx}$ and not $\lim_{n\to \infty}\frac{1}{2}\frac{1}{n} \sum_{k=1}^{n} \ln \left(1+\frac{k}{n}-frac{1}{n}right=\frac{1}{2}\int_0^1{ln(1+x)dx}$ – DatBoi Aug 19 '20 at 14:56
  • Let us continue this discussion in chat. – DatBoi Aug 19 '20 at 15:09