Prove that a sequence converge with definition

Question

I've always had a problem when it comes to proving that a sequence is convergent or divergent using the definition. I tried again to do such an exercise but I just can't do it. Here is my exercise:

$$\left\{\frac{3n^2-1}{10n+5n^2}\right\}.$$

Using limits I figured it out that it converges to $\frac{3}{5}$, now when it comes to proving it by definition I choose a $\epsilon > 0$ but I don't know how to find $k_\epsilon$.

Answer 1

Hint: can you show that $\left|a_n - 3/5\right| \leqslant \left|6 /(5n)\right|$? Can you then finish from there, choosing an appropriate value of $K_\varepsilon$? That is, how can you ensure the final expression is always less than the given $\varepsilon$?

Answer 2

What you need to prove is that you can make the difference

$$\left\lvert\frac{3n^2-1}{10n+5n^2}-\frac35\right\rvert $$

'small' ($< \epsilon$) if you choose $n$ 'big enough' ($>n_\epsilon$).

The first step to do so is is to simplify the above so as to get a quotient of polynomials inside the absolute values:

$$\left\lvert\frac{3n^2-1}{10n+5n^2}-\frac35\right\rvert = \left\lvert\frac{3n^2-1}{10n+5n^2}-\frac{3(2n+n^2)}{5(2n+n^2)}\right\rvert = \left\lvert\frac{3n^2-1}{10n+5n^2}-\frac{6n+3n^2}{10n+5n^2}\right\rvert = \left\lvert\frac{6n-1}{10n+5n^2}\right\rvert.$$

Now the point is to find for each $\epsilon > 0$ an $n_\epsilon \in \mathbb N$ such that

$$\left\lvert\frac{6n-1}{10n+5n^2}\right\rvert < \epsilon\text{, for all }n>n_\epsilon.$$

You now have to work backwards. You should try to find terms $t_1(n),t_2(n),\ldots$ such that

$$\left\lvert\frac{6n-1}{10n+5n^2}\right\rvert \le t_1(n) \le t_2(n) \ldots \le t_k(n)\text{, for all }n,$$

in the hope that the last term $t_k(n)$ is so simple that is it is easy to get from the condition $t_k(n) < \epsilon$ to an equivalent condition $n > r_k(\epsilon)$.

In our example, for $n\ge 1$ we have $\lvert 6n-1\rvert < \lvert 6n\rvert$, so we choose $t_1(n)=\left\lvert\frac{6n}{10n+5n^2}\right\rvert$. So we have

$$\left\lvert\frac{6n-1}{10n+5n^2}\right\rvert \le \left\lvert\frac{6n}{10n+5n^2}\right\rvert$$

That term $t_1(n)$ is already simpler than the original, the enumarator is is monomic polynomial. Similiarly, we can decrease the denominator to increase the quotient: We have $\lvert 5n^2+10n\rvert \ge \lvert 5n^2\rvert$ and hence use $t_2(n)=\left\lvert\frac{6n}{5n^2}\right\rvert$:

$$\left\lvert\frac{6n-1}{10n+5n^2}\right\rvert \le \left\lvert\frac{6n}{10n+5n^2}\right\rvert \le \left\lvert\frac{6n}{5n^2}\right\rvert \tag{1}\label{eq1}.$$

Note that $t_2(n)$ is again simpler than $t_1(n)$. In fact if we cancel one $n$, we get

$$t_2(n)=\left\lvert\frac{6n}{5n^2}\right\rvert=\frac65\left\lvert\frac1{n}\right\rvert = \frac65\frac1{n}.$$

Now that is simple enough to easily revert the comparison with $\epsilon$:

$$\begin{eqnarray} \frac65\frac1{n} & < & \epsilon \Longleftrightarrow \\ \frac65 & < & n\epsilon \Longleftrightarrow \\ \frac65\frac1{\epsilon} & < & n \Longleftrightarrow \\ n & > & \frac6{5\epsilon}\\ \end{eqnarray}$$

So now we know that for $n > \frac6{5\epsilon}$ we have $t_2(n) < \epsilon$. But from $\eqref{eq1}$ this also implies

$$\left\lvert\frac{6n-1}{10n+5n^2}\right\rvert < \epsilon. \tag{2} \label{eq2}$$

What remains to do is just define $n_\epsilon:=\lceil \frac6{5\epsilon} \rceil $, and then $n > n_\epsilon$ implies $n > \frac6{5\epsilon}$ and that implies $\eqref{eq2}$ and that is what we want.