This question is similar to If $z_n \to z$ then $(1+z_n/n)^n \to e^z$
In this case $z_n$ does not converge but $ \frac{z_n}{n} \rightarrow 0 $.
Asked
Active
Viewed 101 times
0
-
Of course not. Let $(z_n)$ be the sequence $0,-100, 0, 0, 0, \dots.$ – zhw. Apr 28 '17 at 16:10
1 Answers
0
Consider $$\ln\left[(1+z_n/n)^n\right] =n\ln(1+z_n/n)=z_n-\frac{z_n^2}{2n}+\cdots.$$ This will usually be less than $z_n$ eventually.
Angina Seng
- 158,341
-
Does that eventually happen at $n = 2 $ for $z_n = \log(n)$? I understand when $ z_n = \sqrt{n}$, then it will happen at $n = \inf$ – user3727929 Apr 28 '17 at 15:59