3
$\begingroup$

Could someone help me through this problem? Prove that $$\lim_{n \to{+}\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)=0$$

  • 1
    I think you have a typo: it should rather be $n\to \infty$ and not $x\to \infty$, otherwise the result doesn't hold.2012-04-24
  • 0
    do you mean $\displaystyle\lim_{n \to{+}\infty}$?2012-04-24
  • 0
    I presume you mean $n \rightarrow \infty$. First try showing $\lim_{n \rightarrow \infty} \frac{1}{n} = 0$.2012-04-24

3 Answers 3

2

you have $$\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$$ and $$0\leq \frac{1}{n(n+1)}\leq \frac{1}{n^2}$$ then because $$\lim_{n \to +\infty} \frac{1}{n^2}=0$$

we have $\lim_{n \to +\infty}(\frac{1}{n}-\frac{1}{n+1})=0$.

  • 1
    Once you have $1\over n(n + 1)$, the problem has fallen; no need to drag $1\over n^2$ into it. IMHO, of course.2012-04-25
5

Hint: convert that difference into a single fraction.

4

Hint First convince yourself that $\displaystyle\lim_{n\to\infty}\frac{1}{n}=0$ and subsequently that $\displaystyle\lim_{n\to\infty}\frac{1}{n+1}=0$.

  • 0
    Just as the OP, you're using $x$ and $n$ simultaneously when you should either choose $n$ or $x$.2012-04-24
  • 0
    @PeterTamaroff Silly mistake, fixed now2012-04-24