Prove that
$$\lim_{n\to\infty} \frac{n^z n!}{z(z+1)\cdots (z+n)}=\frac{1}{z}\prod_{n=1}^\infty \frac{(1+\frac{1}{n})^z}{1+\frac{z}{n}}$$?
$$\lim_{n\to\infty} \frac{n^z n!}{z(z+1)\cdots (z+n)}=\frac{1}{z}\prod_{n=1}^\infty \frac{(1+\frac{1}{n})^z}{1+\frac{z}{n}}$$?
Solution:
To prove that
$\large \lim_{n \to \infty}\frac{\prod_{k=1}^n(1+\frac{1}{k})}{n}=1$
expand out the numerator, which will result in
$\prod_{k=1}^n(1+\frac{1}{k})=\prod_{k=1}^n(\frac{k+1}{k})$
writing this out explicitly we have
$\frac{2}{1}\times \frac{3}{2} \cdots \times \frac{n}{n-1} \times \frac{n+1}{n}$
notice that the numerator of the first fraction, all the intermediate fractions, and the denominator of the last fraction cancels out, leaving $n+1$
Thus
$\large \lim_{n \to \infty}\frac{\prod_{k=1}^n(1+\frac{1}{k})}{n}=\lim_{n \to \infty}\frac{n+1}{n}\rightarrow 1$
$\large \lim_{n \to \infty}\frac{\prod_{k=1}^n(1+\frac{1}{k})}{n}=1$
expand out the numerator, which will result in
$\prod_{k=1}^n(1+\frac{1}{k})=\prod_{k=1}^n(\frac{k+1}{k})$
writing this out explicitly we have
$\frac{2}{1}\times \frac{3}{2} \cdots \times \frac{n}{n-1} \times \frac{n+1}{n}$
notice that the numerator of the first fraction, all the intermediate fractions, and the denominator of the last fraction cancels out, leaving $n+1$
Thus
$\large \lim_{n \to \infty}\frac{\prod_{k=1}^n(1+\frac{1}{k})}{n}=\lim_{n \to \infty}\frac{n+1}{n}\rightarrow 1$
No comments:
Post a Comment