question on Natural Log, $\lim \limits_{n\to∞ }(1+\frac{1}{n} + \frac{1}{n^2})^n $

Question

I'm curious what is the solution of this. Is this just same as ordinary natural log?

$\lim \limits_{n\to∞ }(1+\frac{1}{n} + \frac{1}{n^2})^n =e?$

A few people says the $\frac{1}{n^2}$ just goes to $'0'$, so it's same as $'e'$. But why? Why $\frac{1}{n}$ remains meaningful, while $\frac{1}{n^2}$ goes to zero??

I'm asking this because I'm stuck while deriving a formula. I post the part of deriving as a picture. Look. From 1st, they go to 3rd line, by using 2nd. And I assume this suppose $\lim \limits_{n\to∞ }(1+\frac{1}{n} + \frac{1}{n^2})^n $ is just same as 'e'.

Thank you genius

Answer 1

Hint: For any $\epsilon >0$ we have $1+\frac 1 n \leq 1+\frac 1 n+\frac 1 {n^{2}} \leq 1+\frac {1+\epsilon} n$ for $n$ sufficiently large. Apply logarithm, multiply by $n$ and take the limit.

Answer 2

Remember that $(1+\epsilon)^n\simeq 1+\epsilon n$ for $\epsilon\to 0$. If $\epsilon\sim O(1/n)$ then $\epsilon n \sim O(1)$ so such a term is relevant. If $\epsilon \sim O(1/n^2)$ then $\epsilon n\sim O(1/n)\to 0$.

Hence if you have $\left(1 + \frac{1}{n}+\frac{1}{n^2}\right)$ you have something in the form $(1+\epsilon_1 +\epsilon_2)^n$ with $\epsilon_1\sim O(1/n)$ and $\epsilon_2 \sim O(1/n^2)$. So you'll have $1+\epsilon_1 n + \epsilon_2 n$. The term in $\epsilon_1$ is of order $1$ (actually it's $e-1$) and the other is of order $1/n$ and becomes negligible.

To see things more rigorously you can take the log... $$\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n = e^{n\log(1+1/n+1/n^2)}=e^{(1+O(1/n)) +(1/n+O(1/n^2))}=e^{1+O(1/n)}\to e$$

Answer 3

$$\begin{align} \lim_{n\to \infty} \left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n &=e^{\lim_{n\to\infty} n\left(\frac{1}{n}+\frac{1}{n^2}\right)}\\ &=e^{\lim_{n\to\infty} \left(1+\frac{1}{n}\right)} \\ &=e \end{align}$$

Answer 4

$$\left(1+\frac{n+1}{n^2} \right)^{n}=\left(1+\frac{n+1}{n^2} \right)^{n \cdot\frac{n}{n+1} \cdot \frac{n+1}{n}}=\left(\left(1+\frac{1}{y_n} \right)^{y_n}\right)^{1+\frac{1}{n}} \rightarrow e $$

Answer 5

$$\left(1+\frac{1}{n}+\frac{1}{n^2}\right)=\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n^2+n}\right) $$ so if $\left(1+\frac{1}{n}\right)^n\to e$, then $\left(1+\frac{1}{n^2+n}\right)^n\to 1$ and $\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n\to e$.

Answer 6

When $n>1$, we have $$\left(1+\frac{1}{n}\right)^n<\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n <\left(1+\frac{1}{n-1}\right)^n =\left(1+\frac{1}{n-1}\right)^{n-1}\cdot\left(1+\frac{1}{n-1}\right).$$