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

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

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    $\begingroup$ 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. $\endgroup$ – Kaz Apr 25 '12 at 0:20
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Hint: convert that difference into a single fraction.

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

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  • $\begingroup$ Just as the OP, you're using $x$ and $n$ simultaneously when you should either choose $n$ or $x$. $\endgroup$ – Pedro Tamaroff Apr 24 '12 at 22:44
  • $\begingroup$ @PeterTamaroff Silly mistake, fixed now $\endgroup$ – Milosz Wielondek Apr 24 '12 at 22:45

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