Evaluating a limit without squeeze principle

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

Thanks.

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Why do you want not to use this argument? –  Seirios Jan 22 '13 at 16:53
This is pretty easy to do from the definition of limit with a $\epsilon-\delta$ argument... –  Jonathan Christensen Jan 22 '13 at 16:53
I'm in a beginning calculus class. the question is in a section where the squeeze principle hasn't been introduced. –  Gorg Jan 22 '13 at 16:56
First guess the answer. Then, as @JonathanChristensen wrote, try using an $\epsilon$-$\delta$ argument. Let $\epsilon>0$, choose $N$ such that $\frac{1}{N}<\epsilon$, etc... –  copper.hat Jan 22 '13 at 16:58
If you want to evaluate without formal proof, think of a huge $n$, and decide roughly where $(-1)^n/(n+1)$ is. –  André Nicolas Jan 22 '13 at 17:24

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 $x$.

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Thanks very much. –  Gorg Jan 23 '13 at 20:05