Finding and proving $\bigcap\limits_{n=1}^{\infty} \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$ Letting $A_n = \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$, I am trying to find $\bigcap\limits_{n=1}^{\infty} A_n$. I believe the answer is $\left[0, \frac{1}{3}\right)$. I don't have any intuition for why this is the case other than attempting to prove that the answer is $\left[0, \frac{1}{2}\right)$. Approacing the left and right endpoints as sequences and taking $n \to \infty$ didn't help.
I'm going to attempt to prove this.
Let $x \in \left[0, \frac{1}{3}\right)$. Then $0 \leq x < \frac{1}{3}$. We must show that $x \in A_n$ for all $n$. So fix $n \geq 1$. Certainly $- \frac{1}{n} < 0 \leq x$. Now we show that $x < \frac{n}{2n+1}$. Since $n \geq 1$, $2n \geq 2$ and $2n + 1 \geq 3$, so $\frac{1}{2n+1} \leq \frac{1}{3}$. Certainly, because $n \geq 1$, $\frac{n}{2n+1} \geq \frac{1}{2n+1}$, but that doesn't help, so at this point I'm stuck.
For the reverse direction, let $x \in \bigcap\limits_{n=1}^{\infty} A_n$, so $x \in \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$ for all $n$. So $- \frac{1}{n} < x < \frac{n}{2n+1}$. Outside of taking $n \to \infty$, I don't know how to proceed.
Any help or hints would help.
 A: Your answer of $[0,\frac 13)$ is correct.  I would approach it by computing the intervals for $n$ from $1$ to $10$ first to see what I learn.  The $-\frac 1n$ end it converging toward $0$ but always less, so I would guess that the lower bound is $0$ and it is included.  Now to prove it, note that $0 \gt -\frac 1n$ for all $n$, so is included.  Any negative number is less than $-\frac 1n$ for large enough $n$, so is not included.  The top end is increasing with $n$ so we are limited by the $n=1$ case, which is $\frac 13$.  We can't have any more than is included in $A_1$, which is up to but not including $\frac 13$, so that it the upper limit.  You just need to prove that for any $x \lt \frac 13, x \lt \frac n{2n+1}$ to be done.  Showing that $\frac n{2n+1}$ is increasing is sufficient.
A: Let $A_n = \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$. We want to find $\bigcap_{n=1}^\infty A_n = \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$.

*

*Suppose $x\in \bigcap_{n=1}^\infty A_n$. Then, $x\in A_n$ for all $n\ge 1$. That is, $x > -\frac1n$ for all $n\ge 1$, and taking limits as $n\to\infty$ gives you $x\ge  0$. Also, $x < \frac{n}{2n+1}$ for all $n\ge 1$. This is where you have to be careful - simply taking limits as $n\to\infty$ will not help. Here's why:
$$x < \frac{n}{2n+1} = \frac12\cdot \left(1  - \frac{1}{2n+1} \right) \quad (\forall n\ge 1)$$
and $\frac12\cdot \left(1  - \frac{1}{2n+1} \right)$ is an increasing function of $n$. As a result, if $x < \frac12\cdot \left(1  - \frac{1}{2k+1} \right)$ for some $k\in \mathbb N$, then we certainly have $x < \frac12\cdot \left(1  - \frac{1}{2m+1} \right)$ for all $m \ge k$. Thus, put $n=1$ and see that $x < \frac13$. As discussed, $x < \frac12\cdot \left(1  - \frac{1}{2n+1} \right)$ for all $n\ge 1$ is then automatically satisfied. Hence, $x \in [0, \frac13)$.


*On the other hand, suppose $x\in [0,\frac13)$. You have already shown $x > -\frac1n$ for all $n\ge 1$. $\frac12\cdot \left(1  - \frac{1}{2n+1} \right)$ is an increasing function of $n \in \mathbb N$, and its minimum (over natural numbers) is attained at $n=1$. In particular, this minimum is $\frac13$. Thus $x < \frac13$ ensures that $x < \frac12\cdot \left(1  - \frac{1}{2n+1} \right)$ for all $n\ge 1$. Consequently, $x\in \bigcap_{n=1}^\infty A_n$, as desired.
Let me know if you have any questions.
