Evaluating a limit without squeeze principle

Question

Please, I need help to evaluate $$\lim_{n\to \infty} \frac{(-1)^n}{n+1}$$ without using the squeeze principle.

Thanks.

Answer 1

This is a direct application of the definition of limit; I’ll take you through this one in detail, so you that you can use it as a model for other, possibly harder problems.

I expect that you can easily guess that the limit is $0$, so it’s just a matter of proving this to be the case. The definition says that the limit is $0$ if and only if

$\qquad$ for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $\left|\frac{(-1)^n}{n+1}-0\right|<\epsilon$ whenever $n\ge m_\epsilon$.

Of course $\left|\frac{(-1)^n}{n+1}-0\right|=\frac1{n+1}$, so we may translate the definition to the following slightly simpler form: $0$ is the limit of the sequence if and only if

$\qquad\qquad$ for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $\frac1{n+1}<\epsilon$ whenever $n\ge m_\epsilon$.

What does it take to make $\frac1{n+1}<\epsilon$? we need to have $1<(n+1)\epsilon$, or $n+1>\frac1\epsilon$. (Both $\epsilon$ and $n+1$ are positive, so multiplying and dividing by them don’t change the directions of the inequality.) In other words, if we guarantee that $n+1>\frac1\epsilon$, we’ll know that $\frac1{n+1}<\epsilon$. If we let $m_\epsilon$ be any integer that is at least $\frac1\epsilon$, we’ll be in business: if $n\ge m_\epsilon$, then $n+1>m_\epsilon\ge\frac1\epsilon$, and therefore

$$\left|\frac{(-1)^n}{n+1}-0\right|=\frac1{n+1}<\epsilon\;,$$

exactly as we wanted. If you want to write down a specific choice of $m_\epsilon$, you can let

$$m_\epsilon=1+\left\lfloor\frac1\epsilon\right\rfloor\;,$$

where $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$.