$\left\{\frac{1}{n}\right\}$ is a Cauchy sequence. Determine $N_0$ such that $|u_n - u_m| < 0.021$, whenever $m, n > N_0$.
a. $48$
b. $45$
c. $46$
d. $47$
The correct answer for this was $48$. But the way I solved it, I have to divide my answer by $2$ in order to get it. I don't understand where I did wrong. Can someone please tell me how to solve this?
Edit: This is how I got to my answer:
\begin{align*} |u_m - u_n| &\le \frac{1}{m} + \frac{1}{n} &\text{(Triangle inequality)} \\ |u_m - u_n| &< \frac{2}{N_0} < 0.021 \\ N_0 &> \frac{2}{0.0021} \\ N_0 &> 95.238 \end{align*}
The first thing you need to know is that your solution of $96$ would be perfectly fine, if it weren't for the multiple choice. If $N_0 = 96$, then your proof (with a little tidying) would show that $$m, n > N_0 \implies \left|\frac{1}{m} - \frac{1}{n}\right| < 0.021,$$ which is exactly what you want. Indeed, any number greater than $96$ (or any other $N_0$ that works), because if $m$ and $n$ are greater than some $N_1$ larger than $96$, then they are definitely larger than $96$, and the result still follows. So, user10354138's (now deleted) sarcastic comment claiming that $N_0 = \operatorname{TREE}(3)$ is perfectly valid too ($\operatorname{TREE}(3)$ is a famously, almost indescribably, big natural number that is certainly larger than $96$).
But, since $96$ is not an option, we should probably look for smaller numbers that $N_0$ could be. While you very naturally found the $96$ number, that isn't to say there isn't smaller possible numbers that will work! We just might need to use a different argument to get a tighter bound.
One thing to note here is that all of the points in the sequence are positive. This implies that, when computing $|u_m - u_n|$, the result will never be the same as $|u_n| + |u_m|$. Using the triangle inequality as you did will give you a valid inequality, but it won't be sharp. You can only get $|u_n - u_m| = |u_n| + |u_m|$ when $u_n$ and $u_m$ have opposite signs (e.g. $|-3 - 1| = |-3| + |1|$), but you can never get this equality when they are the same sign! So, while your argument is valid, this step does introduce some slack.
Instead, consider, since $u_n > 0$ for all $n$, that $|u_n - u_m| < u_n$ if $u_n$ is larger than $u_m$. This is because, if $u_n > u_m$, then $u_n - u_m > 0$, so $|u_n - u_m| = u_n - u_m$. But, since $u_m > 0$, we know $u_n - u_m < u_n$.
If $u_m$ is larger, then $|u_n - u_m| = |u_m - u_n| < u_m$ for the same reason. Thus, we can express this like so: $$|u_n - u_m| < \max\{u_n, u_m\}.$$ Again: this only holds because $u_m, u_n$ are positive!
So, if we can find $N_0$ such that $n > N_0 \implies u_n < 0.021$, then \begin{align*} m, n > N_0 &\implies u_n < 0.021 \text{ and } u_m < 0.021 \\ &\implies \max\{u_n, u_m\} < 0.021 \\ &\implies |u_n - u_m| < 0.021. \end{align*} What $N_0$ makes $u_n < 0.021$? We can just take $N_0 = \frac{1}{0.021}$, which rounds up to $48$. This indicates the answer is (a).
I can even go further: $48$ is the least possible value for $N_0$. This is because, as we take larger and larger $m$, we get $$|u_n - u_m| \to |u_n - 0| = u_n,$$ so if we let $n < 48$, i.e. $n \le 47$ i.e. $$u_n \ge \frac{1}{47} = 0.\overline{0212765957446808510638297872340425531914893617},$$ then for large $m$, say $m > 1000$, we would have $u_m < 0.001$, and so $$|u_n - u_m| > 0.02127 - 0.001 = 0.02117 > 0.021.$$ So, even $N_0 = 47$ (or anything lower) will be insufficient!
That's a more meta-gamey way of picking the right answer. If any of the lower choices for $N_0$ would work, then so would the higher choice. So, if (d) were a correct choice, then (a) would be too. If (b) were correct, then all of them would be correct! In this way, without any calculation, you could just pick the largest number (a), and assuming any of the answers are correct, (a) would be one of them.