limit of rational sequence is not-necessarily rational and another limit question 1)Disprove that if $a_n$ is rational for all $n$ then its limit is rational. The first sequence that came to my mind is $a_n = ((n+1)/n)^n$, such that every element $a_n$ is a proportion of two  integers in some power $n$, so it is a rational number, however its limit is $e$. Is this legitimate proof?
2) How do I rigorously prove that $\lim (1+1/{n^2})^n = 1$ and $\lim(1+1/n)^{n^2}=\infty$. It is clear intuitively as $n=O(n^2)$, but how do I show it in a more constructive manner?
 A: 2) As you probably know, $\displaystyle \lim_{n \rightarrow \infty}\left(1+\frac{a}{n} \right)^n=e^a$. For any $a>0$, the following inequalities hold for $n>1/a$,
$$1\leq \left(1+\frac{1}{n^2} \right)^n \leq \left(1+\frac{a}{n} \right)^n$$
hence, as $n$ goes to infinity, we get 
$$1\leq \liminf_{n \rightarrow \infty}\left(1+\frac{1}{n^2} \right)^n\leq\limsup_{n \rightarrow \infty}\left(1+\frac{1}{n^2} \right)^n\leq \lim_{n \rightarrow \infty}\left(1+\frac{a}{n} \right)^n=e^a$$
Since $a$ is an arbitrary positive number and $\inf_{a>0} e^a=1$, then the limit $\displaystyle \lim_{n \rightarrow \infty}\left(1+\frac{1}{n^2} \right)^n$ exists and it is equal to $1$.
You can prove the other limit in a similar way.
P.S. If you know that $\ln(1+x)\leq x$ for $x>0$ then you can also make it in this way:
$$1\leftarrow 1\leq \left(1+\frac{1}{n^2} \right)^n=\exp\left(n\ln\left(1+\frac{1}{n^2} \right)\right)\leq \exp\left(\frac{1}{n} \right)\rightarrow 1.$$
A: 1) Yes, if $a_n=\left(1+\frac1n \right)^n$, then $a_n \in \mathbb Q (\forall n \in \mathbb N)$ and
$$\lim_{n \rightarrow \infty} a_n=e \not \in \mathbb Q$$
Example else $\left(b_n \right):$
$$b_1=1; b_2=1.4; b_3=1.41; b_4=1.414; b_5=1.4142; b_6=1.41421;...$$
Where $b_n$ gives the first $n$ digits of $\sqrt2$
Then 
$$\lim_{n \rightarrow \infty} b_n=\sqrt2 \not \in \mathbb Q$$
A: It is easy to give examples of sequences which take rational values but have irrational limit (in fact Cantor constructed the real number system as limits of such convergent sequences of rational numbers). My favorite non-trivial example is the sequence $$x_{1} = 1, x_{n + 1} = \frac{1}{2}\left(x_{n} + \frac{2}{x_{n}}\right)$$ which by definition takes rational values only and converges to $\sqrt{2}$.

Since you already know that the limit of $a_{n} = (1 + (1/n))^{n}$ is $e$, it follows that $b_{n} = (1 + (1/n^{2}))^{n^{2}}$ also tends to $e$ and hence $c_{n} = (1 + (1/n^{2}))^{n} = b_{n}^{1/n} \to 1$. Using similar technique and noting and $e > 1$ you can prove that $d_{n} = (1 + (1/n))^{n^{2}} \to \infty$. Note that the above arguments are based on the following two limits:


*

*If $x > 0$ then $x^{1/n} \to 1$ as $n \to \infty$.

*If $x > 1$ then $x^{n} \to \infty$ as $n \to \infty$.


I hope you are aware of these simple results (including their proofs).

In case you don't want to use any information about $e$ then you can proceed as follows. By the Bernoulli's Inequality we have $$1 > \left(1 - \frac{1}{n^{2}}\right)^{n} > 1 - \frac{1}{n}$$ and hence on taking limit as $n \to \infty$ and using squeeze theorem we get $$\lim_{n \to \infty}\left(1 - \frac{1}{n^{2}}\right)^{n} = 1\tag{1}$$ Similarly we can prove $$\lim_{n \to \infty}\left(1 - \frac{1}{n^{4}}\right)^{n} = 1\tag{2}$$ and dividing $(2)$ by $(1)$ we get $$\lim_{n \to \infty}\left(1 + \frac{1}{n^{2}}\right)^{n} = 1\tag{3}$$ Next using Bernoulli's Inequality again we get $$\left(1 + \frac{1}{n}\right)^{n^{2}} > 1 + n$$ and therefore $$\left(1 + \frac{1}{n}\right)^{n^{2}} \to \infty$$ as $n \to \infty$.
