Find the limit of $a_n$ from, $n=1$ to $\infty$, in the following cases, and justify your answers using the definition of convergence. (Note the definition of convergence I'm using: a sequence an converges to a real limit if given any $\epsilon > 0$, there exists an $N$ belonging to the natural numbers such that: $n \ge N \Rightarrow |a_n - a| < \epsilon$).

i)$$a_n = \frac{2}{n} + \frac{3}{n^2}$$

This is a homework but I have solutions but am unable to understand all the steps. Solution: Guess limit $a = 0$. Fix $\epsilon > 0$. Find a simpler sequence which has bigger terms than $a_n$ and converges to zero. $$an = \frac{2}{n} + \frac{3}{n^2} < \frac{2}{n} + \frac{3}{n} = \frac{5}{n}$$

So take $N > \frac{5}{\epsilon}$, $n \ge N$ (from definition of convergence) $$\Rightarrow n > \frac{5}{\epsilon} \Rightarrow \frac{5}{\epsilon} < \epsilon \Rightarrow n > a_n$$.

The parts of the solution I don't understand is why to take $N$ as $\displaystyle >\frac{5}{\epsilon}$. Also how to get from $\displaystyle \frac{5}{\epsilon}< \epsilon$ to $n>a_n$ and how showing that $n>a_n$ proves it converges to $0$. Thanks for any help

  • $\begingroup$ please try to use latex next time ;) $\endgroup$
    – Ant
    Jan 6 '14 at 14:33

$$a_n = \frac{2}{n} + \frac{3}{n^2}$$

What you want to prove is that

$$\forall \epsilon > 0 \ \ \exists N: n \ge N \Rightarrow a_n < \epsilon$$

(note we can drop the absolute value since $a_n > 0$)

Now, we know that $$a_n < \frac{5}{n} (1)$$

If we take $N = \frac{5}{\epsilon}$, then for $n > N = \frac{5}{n}$ we have $$a_n < \frac{5}{\frac{5}{\epsilon}} = \epsilon$$, (just substitute in (1))

That is $$a_n < \epsilon$$ for $n > N = \frac{5}{\epsilon}$ as we wanted to show


What you want as a consequence of $n>N$ is:

$\dfrac{5}{n}<\varepsilon$ or equivalently:


This is exactly what you get if you choose for $N=\dfrac{5}{\varepsilon}$. Then $n>N\iff n>\dfrac{5}{\varepsilon}\iff\dfrac{5}{n}<\varepsilon$.

In fact it is only of importance that $n>N$ implies $\dfrac{5}{n}<\varepsilon$ so the arrows backwards are not relevant.

To find a suitable $N$ in general start looking at the equation $a_{n}<\varepsilon$ and try it to 'deform' into something like $n>N\left(\varepsilon\right)$.

Again: it is enough allready if $n>N\left(\varepsilon\right)$ implies $a_{n}<\varepsilon$.


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