Im stuck with the ε-δ definition of a limit of succession

So, I have this problem $$ \lim_{n\to\infty} \frac{1}{n} +\frac{(-1)^n}{n^2} = 0$$ that i have to resolve with the definition of the limit of succession. I have done this $ \lvert \frac{1}{n} +\frac{(-1)^n}{n^2} - 0\rvert \lt \varepsilon$ $$\lvert\frac{1}{n} +\frac{(-1)^n}{n^2}\rvert \lt \varepsilon $$ $$ \lvert \frac{n+(-1)^n}{n^2} \rvert \lt \varepsilon$$ $$\frac{\lvert{n+(-1)^n}\rvert} {\lvert{n^2}\rvert}\lt \varepsilon $$ $$ \frac {\lvert n\rvert + \lvert(-1)^n\rvert}{\lvert{n^2}\rvert}\lt \varepsilon $$ but i dont know how i could resolve this problem with the $\lvert (-1)^n \rvert $ Can anyone explain me this problem?

Solution 1:

As far as I know usually there is no $\delta$ in the definition of a sequence, but $\varepsilon$ and $N$ is used. But of course you can call the variables in a mathematical statement as you like.

We want to show that $$\lim_{n\to\infty}( \frac{1}{n} +\frac{(-1)^n}{n^2}),\tag 1$$ so we have to show that for each $\varepsilon>0$ there exists an $N_\varepsilon$ such that $$\left| \frac{1}{n} +\frac{(-1)^n}{n^2}\right|<\varepsilon,\quad \forall n\in \mathbb N:\;n>N_\varepsilon\tag 2$$

We use $N$ with the subscript $\varepsilon$ to make clear that $N$ depends on $\varepsilon.$ Different values of $\varepsilon$ need different values of $N.$

So given $\varepsilon$, we want to calculate $N_\varepsilon$ such that $(2)$ holds. But calculating tusch an $N_\varepsilon$ directly from $(2)$ is difficult because the LHS (left hand side) of the inequality $(2)$ is a complicated expression. But we can help us. We do not need the least possible $N_\varepsilon$. So we can substitute the LHS of $(2)$ by an expression that is greater than $\left| \frac{1}{n} +\frac{(-1)^n}{n^2}\right|$. If this greater expression is less than $\varepsilon$ then $\left| \frac{1}{n} +\frac{(-1)^n}{n^2}\right|$ is less than this greater expression and therefore less than $\varepsilon$.

We know that the following hold for real numbers $a,b$:

$$|a+b|\le |a|+|b|$$ $$|ab|=|a||b|$$ and from the latter follows $$|a^n|=|a|^k, \forall k \in \mathbb N$$ and $$\left |\frac a b \right|=\frac{|a|}{|b|}$$

So we have

$$\left| \frac{1}{n} +\frac{(-1)^n}{n^2}\right|\le \left| \frac{1}{n} \right | +\left| \frac{(-1)^n}{n^2}\right|= \frac{|1|}{|n|} + \frac{|(-1)|^n}{\left|n^2\right|} \\ = \frac 1 {|n|} + \frac 1 {|n^2|}=\frac 1 n + \frac 1 {n^2} \le \frac 1 n +\frac 1 n =\frac 2 n \tag 3$$

The last $\le$ hold because $n\le n^2$ and so $\frac 1 {n^2} \le \frac 1 n$. So instead of $(2)$ we now try to solve

$$\frac 2 n \lt \varepsilon, \quad \forall n\in \mathbb N:\;n>N_\varepsilon\tag 4$$

This is easy, we get

$$n>\frac 2 \varepsilon \tag5$$

So whenever $(5)$ holds, $(4)$ holds, too. And because of

$$\left| \frac{1}{n} +\frac{(-1)^n}{n^2}\right|\le \frac 2 n ,$$

that follows by $(3)$, $(2)$ holds, too. So because of $(5)$ we choose $$N_\varepsilon=\frac 2 \varepsilon$$ and then $(2)$ holds.