# Show that if $\{x_n\}$ converge and $\{y_n\}$ diverge then $\{ax_n + by_n\}$ diverge, for $b \ne 0$

Let $$\{x_n\}$$ be a convergent sequence and $$\{y_n\}$$ a divergent sequence. Prove that $$\{ax_n + by_n\}$$ diverges for $$b\ne 0$$.

Intuitively this is obvious to me, however i want a formal proof.

Using the definition of a limit we know that:

$$\lim_{n\to \infty}ax_n = A \stackrel{\text{def}}{\iff} \{\forall \varepsilon > 0\ \exists N \in \mathbb N : \forall n \ge N \implies |ax_n - A| < \varepsilon\}$$

On the other hand we know that $$y_n$$ diverges, thus: $$\lim_{n \to \infty}by_n = \exists! \stackrel{\text{def}}{\iff} \{\exists \varepsilon >0 \ N \in \mathbb N : \exists n \ge N \implies |by_n - B| \ge \varepsilon\}$$

Choose some $$\varepsilon$$: $$\begin{cases} |ax_n - A| < {\varepsilon \over 2} \\ |by_n - B| \ge {\varepsilon \over 2} \end{cases}$$ or: $$\begin{cases} |ax_n - A| < {\varepsilon \over 2} \\ -|by_n - B| \le -{\varepsilon \over 2} \end{cases}$$

This is where I got stuck. It feels like i have to use some sort of triangular inequality (either direct or reverse) and find a contradiction. How do I proceed from this point?

## 2 Answers

There's more than one way to skin a cat.

The easiest way to prove your point is by knowing another fact, that is:

If $$\{a_n\}$$ converges and $$\{b_n\}$$ converges and $$\alpha, \beta\in\mathbb R$$, then $$\{\alpha a_n + \beta b_n\}$$ also converges.

Using this, you can easily construct a proof by contradiction. That is, you can, assuming that $$\{x_n\}$$ converges and $$\{ax_n + by_n\}$$ converges, prove that $$\{y_n\}$$ must also converge, since $$y_n = \frac{1}{b}\left(ax_n + by_n\right) + \left(-\frac{a}{b}\right)x_n$$

However, if you insist on going by definitions, then you shouldn't just "choose some $$\epsilon$$." The fact that $$\{y_n\}$$ does not converge gives you a place where you can start to build your $$\epsilon$$. In particular, choosing some $$B$$, you can take the $$\epsilon_y$$ which satisfies the property that $$\forall N\in\mathbb N\exists n>N: |y_n - B|>\epsilon$$.

You can then use this to prove that $$a\cdot A + b\cdot B$$ cannot be a limit, and since every number can be written as $$aA+bB$$ for some value $$B$$, this proves the sequence doesn't converge.

• Why would you want to skin a cat? – user608030 Dec 3 '18 at 13:53
• @ZacharySelk Presumably because it got out of the bag, and proceeded to get my tounge in a cat-and-mouse game with some other fat cat, and now it's looking at me like the cat that ate the canary. – 5xum Dec 3 '18 at 14:02
• Thank you very much for your answer, that was very helpful. – roman Dec 3 '18 at 14:55

For the sake of completeness I'm putting here my further steps based on the hints.

Let: $$\lim_{n\to \infty}x_n = A\\ \lim_{n\to \infty}y_n = B\\$$

We want to show that $$\lim_{n\to\infty}(ax_n + by_n)$$ exists. Suppose $$x_n$$ and $$y_n$$ converge. It is necessary and sufficient $$(1)$$ to show that:

$$x_n = A + \alpha_n\\ y_n = B + \beta_n$$

Where $$\alpha_n$$ and $$\beta_n$$ are infinitely small sequences. Then by definition of a limit we have: $$|x_n - A| < \varepsilon \iff |\alpha_n| < \varepsilon \iff \lim_{n\to\infty}\alpha_n = 0 \\ |y_n - B| < \varepsilon \iff |\beta_n| < \varepsilon \iff \lim_{n\to\infty}\beta_n = 0$$

Take some $$a$$ and $$b$$ and consider the following limit: $$\lim_{n\to\infty}(ax_n + by_n) = \lim_{n\to\infty}\left((aA + bB) + (a\alpha_n + b\beta_n)\right) = aA + bB + \lim_{n\to\infty}(a\alpha_n + b\beta_n)$$

But any linear combination of infinitely small sequences is an infinitely small sequence. So:

$$\lim_{n\to\infty}(ax_n + by_n) = aA + bB$$

Now consider the case when $$y_n$$ diverges, that means that: $$|y_n - B| \ge \varepsilon \iff |\beta_n| \ge \varepsilon \iff \lim_{n\to\infty}\beta_n = \exists !$$

So since $$y_n$$ may not be presented as a sum of some constant and and infinitely small sequence (which violates $$(1)$$) then $$(ax_n + by_n)$$ must also diverge.