0
$\begingroup$

I claim that the sequence $\{a_n\}_{n = 1}^{\infty}$ with $a_n = \dfrac{1}{n(n+1)}$ converges to $0$. Now using this definition of convergence of a sequence

A sequence $\{a_n\}$ converges to a limit $L$ $$\lim_{n\to \infty} = L$$ if, for any $\epsilon > 0$, there exists an $N$ such that $\mid a_n - L \mid < \epsilon$ for $n > N$

$$ \left\lvert \frac{1}{n(n+1)} \right\lvert < \epsilon $$ $$\frac{1}{n(n+1)} < \epsilon$$ $$n(n+1) > \frac{1}{\epsilon}$$ $$n^2 + n > \frac{1}{\epsilon}$$ I am not able to proceed further. How can I find an $N$ not in terms of $n$ that satisfies the definition? I tried dividing it into two cases: $\epsilon \ge 1$ and $\epsilon < 1$ but still couldn't solve it. Is my claim wrong (I don't think so) or am I not able to manipulate the inequality as needed?

  • 6
    You don't need the "smallest possible $N$". Just let $N > \frac{1}{\epsilon}$. Then $n^2 + n > n > N > \frac{1}{\epsilon}$. – 2017-01-02
  • 1
    @SteamyRoot is correct. The following might also be useful for you in the future. If you know that $x_{n}\rightarrow x$ and $y_{n}\rightarrow y$, then $x_{n}y_{n}\rightarrow xy$. In this case, you could take $x_{n}=1/n$ and $y_{n}=1/(n+1)$, both of which you already know go to zero. – 2017-01-02

2 Answers 2

1

You are doing great!

You need to find an $N$ such $n^2+n>\frac{1}{\epsilon}$ for all $n>N$.

Just take $N>\frac{1}{\epsilon}$ (this is possible by the archimidean property.

Now, if $n>N$ we have:

$n^2+n>n>N>\frac{1}{\epsilon}$ as desired.

0

It would have been enough to say the following. Given $\epsilon > 0$, choose an integer $N > \frac{1}{\epsilon}$. For all $n > N$,

$$\left\lvert \frac{1}{n(n+1)} - 0\right\rvert = \frac{1}{n(n+1)} < \frac{1}{n} < \frac{1}{N} < \epsilon.$$