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?


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.

  • 2
    $\begingroup$ Why would you want to skin a cat? $\endgroup$ – user608030 Dec 3 '18 at 13:53
  • 1
    $\begingroup$ @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. $\endgroup$ – 5xum Dec 3 '18 at 14:02
  • $\begingroup$ Thank you very much for your answer, that was very helpful. $\endgroup$ – 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.


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