Prove that $\lim\limits_{n \to{+}\infty}{(\frac{1}{n}-\frac{1}{n+1})}=0$

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

• I think you have a typo: it should rather be $n\to \infty$ and not $x\to \infty$, otherwise the result doesn't hold. – Kuku Apr 24 '12 at 22:38
• do you mean $\displaystyle\lim_{n \to{+}\infty}$? – Milosz Wielondek Apr 24 '12 at 22:38
• I presume you mean $n \rightarrow \infty$. First try showing $\lim_{n \rightarrow \infty} \frac{1}{n} = 0$. – copper.hat Apr 24 '12 at 22:39

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 $\lim_{n \to +\infty}(\frac{1}{n}-\frac{1}{n+1})=0$.
• 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. – Kaz Apr 25 '12 at 0:20
Hint First convince yourself that $\displaystyle\lim_{n\to\infty}\frac{1}{n}=0$ and subsequently that $\displaystyle\lim_{n\to\infty}\frac{1}{n+1}=0$.
• Just as the OP, you're using $x$ and $n$ simultaneously when you should either choose $n$ or $x$. – Pedro Tamaroff Apr 24 '12 at 22:44