Determine $n_0 \in \mathbb{N}$ s.t. $|a_n| \lt 5 \cdot 10^{-4}$ $\forall \ n \geq n_0$ for the sequence $a_n = \frac{(-1)^n}{n}$

Question

As per the title question:

Determine $n_0 \in \mathbb{N}$ s.t. $|a_n| \lt 5 \cdot 10^{-4}$ $\forall \ n \geq n_0$ for the sequence $a_n = \frac{(-1)^n}{n}$.

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Progress so far:

As sequences are a new topic for me, I am not sure if I have fully understood the task at hand. My guess is that I need to substitute $a_n$ into the provided inequality to caclulate the value for n.

$ \begin{align*} |a_n| &\lt 5 \cdot 10^{-4} \\ |\frac{(-1)^{n_{0}}}{n_0}| &\lt 5 \cdot 10^{-4} \\ |\frac{1}{n_0}| &\lt 5 \cdot 10^{-4} \implies n_0 = 2000 \\ a_0 &= \frac{(-1)^{2000}}{2000} \end{align*} $

As an additional point, if I have recognized and understood the index notation correctly, then $a_{n_1} = \frac{(-1)^{2001}}{2001}$.

Could someone please verify that I have correctly calculated $n_0$?

Answer 1

$|\frac{1}{n_0}| \lt 5 \cdot 10^{-4} \implies n_0 = 2000$

Actually that does not follow. What follows is.

$|\frac{1}{n_0}| \lt 5 \cdot 10^{-4} \implies n_0 > 2000$

So $n_0$ may be anything bigger than $2000$. Example $n_0$ could equal $2001$. Actually $n_0$ could equal $4,986,821$ for all we know but $2001$ is the smallest possible answer.

So let's say $n_0 = 2001$.

$a_0 = \frac{(-1)^{2000}}{2000}$

Well, no. $a_{2000} = \frac{(-1)^{2000}}{2000}$.

$a_0$ is undefined and doesn't exist (as $a_0 = \frac {(-1)^0}0$ which is undefined and doesn't exist).

"if I have recognized and understood the index notation correctly"

You have not. You are trying to make complicated and convoluted sense of something that doesn't really exist.

It is very strange and very unusual but not technically wrong that the problem used the notation $n_0$ to indicate the natural number being found. There is no indication that $n_0$ is the first of some sequences of natural numbers $\{n_i\}$. And if there is a sequence of $\{n_i\}$ where $n_0$ is the first term, there is no indication as to what that sequence would be or how (if at all) the terms of the sequence are determined.

So any speculation of any value of $n_1$ is pointless. $n_0$ should be treated as a stand alone variable. In fact I'd so in at least $90\%$ of all texts and posts on this forum the convention is to use a capital $N$ (or maybe a capital $M$ or ... whatever).

Even if there is a sequence of $\{n_i\}$ and even if we knew what it was, the indexing of the $n_0, n_1, n_2$ terms need not have anything to do with the indexing of the $a_1, a_2, a_3$ terms. So your worrying about trying to find some $a_0$ term that is related to $n_0$ and then figuring some $a_1$ being the next term is unnecessary and misguided. What's more; it makes no sense and there is no sense that can be made from it.

$a_0$ is undefined and does not exist because $\frac {(-1)^0}{0}$ is undefined and does not exist.

$a_1 = -1$; $a_2 = \frac 12; a_3 = \frac {-1}3; $ and $a_i = \frac {(-1)^i}i$ and so on.

Bearing in mind that $n_0 = 2001$ and that the index of $0$ means absolutely nothing then:

$a_{n_0} =a_{2001} = \frac {-1}{2001}$ and $|a_{2001}|< 5*10^{-4}$.

Likewise $a_{n_0 - 1} = a_{2000} = \frac {1}{2000}$ (and $|a_{2000}|=5*10^{-4})$ and $a_{n_0 + 1} = a_{2002} = \frac 1{2002}$ ($|a_{2002}|< 5*10^{-4}$).

IMO it'd have been better if this entire problem used "$N$" instead of "$n_0$".

Answer 2

The calculation of $n_0$ is correct however for index notation, the correct form is:

$$a_{n_0}=a_{2000}=\frac{(-1)^{n_0}}{n_0}=\frac{1}{2000}$$

and $$a_{n_0+1}=\frac{-1}{2001}$$