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$$f(x)= \lim_{n\rightarrow\infty} \left(\dfrac{n^n(x+n)(x+\dfrac{n}{2})\cdots(x+\dfrac{n}{n})}{n!(x^2+n^2)(x^2+\dfrac{n^2}{2^2})\cdots(x^2+\dfrac{n^2}{n^2})}\right)^{x/n} , \quad x>0$$

How can I represent this limit in a simple form?

I tried that above fomula $\left(\dfrac{\prod\limits_{k=1}^n \left(\dfrac{kx}{n}+1\right)} {\prod\limits_{k=1}^n \left(\dfrac{k^2 x^2}{n^2}+1\right)}\right)^{x/n}$

help me.

pie
  • 4,192
merow
  • 153

3 Answers3

3

HINT:

$$\dfrac{\ln f(x)}x=\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\ln\left(1+x\cdot\dfrac rn\right)-\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\ln\left(1+x^2\left(\dfrac rn\right)^2\right)$$

Now use The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$

lab bhattacharjee
  • 274,582
1

$$\lim_{n\to+\infty}\frac {n^n\prod_{k=1}^n (x+\frac {n}{k})}{n!\prod_{k=1}^n (x^2+\frac {n^2}{k^2})}$$

$$=\lim_{n\to+\infty}\frac {\prod_{k=1}^n (nx+\frac {n^2}{k})}{\prod_{k=1}^n (kx^2+\frac {n^2}{k})} $$

hamam_Abdallah
  • 62,951
0

Cauchy D'Alembert's Law If $a_n$ is a sequence of real numbers and if $\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n} =l$ the $\lim\limits_{n \to \infty} \sqrt[n]{a_n}=l$

$\left(\frac{n^n}{n!}\right)^{\frac{x}{n}}=e^x$ This can be proved via Cauchy D'Alembert's Law.

Now Lets return to the big fraction i.e $\lim\limits_{n\to\infty}\left(\prod\limits_{k=1}^n\frac{x+\frac{n}{k}}{x^2+\frac{n^2}{k^2}}\right)^{\frac{x}{n}}$, Let $a_n(x)= \prod\limits_{k=1}^n\frac{x+\frac{n}{k}}{x^2+\frac{n^2}{k^2}}$. to make the notations less messier I will write $a_n$ instead of $a_n(x)$.

$$b_n:=\frac{a_{n+1}}{a_n}=\frac{x+1}{x^2+1}\prod\limits_{k=1}^n \left[ \frac{\left(x+ \frac{n+1}{k}\right)\left(x^2+\frac{n^2}{k^2}\right) }{\left(x+\frac{n}{k}\right)\left(x^2+ \frac{(n+1)^2}{k^2}\right)}\right]$$ $$=\frac{x+1}{x^2+1}\prod\limits_{k=1}^n \left[1+\frac{1}{kx+n} \right]\left[1+\frac{2n+1}{k^2x^2 +n^2} \right]^{-1}$$

Let $c_n =\prod\limits_{k=1}^n \left[1+\frac{1}{kx+n} \right]$ , $d_n=\prod\limits_{k=1}^n \left[1+\frac{2n+1}{k^2x^2 +n^2} \right] $ so $b_n= \frac{x+1}{x^2+1} \cdot \frac{c_n}{d_n} $.

$$\ln(1+x) = \sum_{m=1}^ \infty \frac{(-1)^{m+1} x^m}{ m} \ \ \ \ \text{ with radius of convergence = 1 }$$

For all $n>3$ $$\ln(c_n ) = \sum\limits_{k=1}^n \ln\left[1+\frac{1}{kx+n} \right]=\sum\limits_{k=1}^n \sum_{m=1}^ \infty \frac{(-1)^{m+1} }{ m (kx+n)^m} $$ $$= \sum\limits_{k=1}^n\frac{1}{kx+n} +\sum\limits_{k=1}^n \sum_{m=2}^ \infty \frac{(-1)^{m+1} }{ m (kx+n)^m} $$

Note that $\left|\sum\limits_{k=1}^n \sum\limits_{m=2}^ \infty \frac{(-1)^{m+1} }{ m (kx+n)^m} \right|< \sum\limits_{m=1}^ \infty \frac{1}{n^m}= \frac{1}{n-1} $

So $\displaystyle \sum_{k=1}^n\left[\frac{1}{kx+n}\right] -\frac{1}{n-1}<\ln(c_n)<\sum_{k=1}^n\left[\frac{1}{kx+n}\right]+\frac{1}{n-1}$ by squeeze theorem $\displaystyle \lim_{n \to \infty}\ln(c_n)= \lim_{n \to \infty} \sum\limits_{k=1}^n\frac{1}{kx+n} $

$$ \lim_{n \to \infty} \sum\limits_{k=1}^n\frac{1}{kx+n} = \lim_{n \to \infty}\frac{1}{n} \sum\limits_{k=1}^n\frac{1}{x\frac{k}{n}+1} = \int_0^1 \frac{1}{1+xt}dt =\frac{\ln(1+x)}{x}$$

so $\displaystyle \lim_{n \to \infty} c_n = (x+1)^{\frac{1}{x}}$

finding $d_n$ is analogous to $c_n$:

$$ \lim_{n \to \infty} d_n = \lim_{n \to \infty} \sum_{k=1}^n \frac{2n}{k^2x^2+n^2}=\lim_{n \to \infty}=\frac{2}{n}\sum_{k=1}^n \frac{1}{\frac{k^2}{n^2}x^2+1}= \int_0^1 \frac{1}{x^2t^2+1}dt = \frac{2\arctan(x)}{x}$$ so $d_n= \exp\left(\frac{2\arctan(x)}{x}\right)$

Since $b_n= \frac{x+1}{x^2+1} \cdot \frac{c_n}{d_n} $ then $$\lim\limits_{n \to \infty} b_n = \frac{x+1}{x^2+1} \cdot \frac{(x+1)^{\frac{1}{x}}}{\exp\left(\frac{2\arctan(x)}{x}\right)}$$

By Cauchy D'Alembert's Law $\lim\limits_{n \to \infty} \sqrt[n]{a_n(x)} = \frac{x+1}{x^2+1} \cdot \frac{(x+1)^{\frac{1}{x}}}{\exp\left(\frac{2\arctan(x)}{x}\right)}$

$$f(x)=e^x\lim_{n \to \infty}a_n^{x/n}= e^x \cdot \left(\frac{x+1}{x^2+1}\right)^x \frac{(x+1)}{e^{2\arctan(x)}} $$

pie
  • 4,192